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Solved Problems on Conservation of Momentum:- Problem 1:- A 4.88-kg object with a speed of 3.14 m/s strikes a steel plate at an angle of 42.0º and rebounds at the same speed and angle as shown in the below figure. What is the change (magnitude and direction) of the linear momentum of the object? Concept:- Momentum of the object (p) is defined as the mass of the object (m) time’s velocity of the object (v). p = mv …… (1) Solution:- The below figure shows a 4.88-kg object with a speed of 31.4 m/s strikes a steel plate at an angle of 42.0^{°} and rebounds at the same speed with same angle which is shown geometrically. The initial momentum of the object is p_{i} when it strikes the steel plate. To find initial momentum p_{i}, substitute p_{i} for p, 4.88-kg for mass m of the object and 31.4 m/s for velocity of the object in the equation p = mv, p_{i} = mv = (4.88 kg) (31.4 m/s) = 153 kg. m/s …… (2) Since the final momentum p_{f} of the object is equal to the initial momentum p_{i} of the object, therefore, p_{f} = 153 kg. m/s …… (3) In the above figure the angle θ will be, θ = 42° + 42° = 84° Using cosine law, the magnitude of change of the linear momentum of the object Δp will be, Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f})cos θ …… (4) To obtain magnitude of linear momentum Δp, substitute 153 kg. m/s for p_{i}, 153 kg. m/s for p_{f} and 84° for the angle θ in the equation Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f}) cos θ, Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f}) cos θ = √(153 kg. m/s)^{2} + (153 kg. m/s)^{2} - 2 (153 kg. m/s)( 153 kg. m/s) cos 84° = 205 kg. m/s From the above observation we conclude that, the change of the linear momentum of the object will be, 205 kg. m/s and also from the figure the direction of change of the linear momentum of the object is perpendicular to the steel plate. ____________________________________________________________ Problem 2:- Below figure shows an approximate representation of force versus time during the collision of a 58-g tennis ball with a wall. The initial velocity of the ball is 32 m/s perpendicular to the wall; it rebounds with the same speed, also perpenducular to the wall. What is the value of F_{max}, the maximum contact force during the collision? Concept:- Momentum of a body (p) is defined as, p = mv Here, m is the mass of the body and v is the velocity of the body. The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval. So, J = Δp = p_{f} - p_{i} Here p_{f} is the final momentum and p_{i} is the initial momentum of the body. Again the impulse is the area under a force-time graph. Solution:- As the tennis ball strikes with the wall having velocity v and then rebounds with same velocity, thus the change in momentum Δp of the tennis ball will be equal to, Δp = (mv) – (-mv) = 2mv To obtain the change in momentum Δp of the ball, substitute 58-g for m and 32 m/s for v in the equation Δp = 2mv, we get, Δp = 2mv = 2 (58-g) (32 m/s) = 2 (58-g×10^{-3} kg/1 g) (32 m/s) = 3.7 kg.m/s To obtain the impulse we have to find out the area under force-time graph for the trapezoid. So, J = F_{max} (2 ms+6 ms)/2 = F_{max} (4 ms) So, F_{max} = J/4 ms = Δp/4 ms To obtain the maximum contact force F_{max} during the collision, substitute 3.7 kg.m/s for Δp in the equation F_{max}= Δp/4 ms, F_{max}= Δp/4 ms = (3.7 kg.m/s) / (4 ms) = (3.7 kg.m/s) / (4 ms×10^{-3} s/1 ms) = 930 kg. m/s^{2} = (930 kg. m/s^{2}) (1 N/1 kg. m/s^{2}) = 930 N From the above observation we conclude that, the maximum contact force F_{max} during the collision would be 930 N. _______________________________________________________________________ Problem 3:- A croquet ball with a mass 0.50 kg is struck by a mallet, receiving the impulse shown in the graph below. What is the ball’s velocity just after the force has become zero? Concept:- The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval. J_{x} = Δp = p_{fx} – p_{ix} Here p_{fx} is the final momentum and p_{ix} is the initial momentum of the body. So, p_{fx} = J_{x} + p_{ix} Momentum p_{x} of a particle is equal to the mass of the particle times velocity v_{x} of the particle. So, p_{x} = m v_{x} Or, v_{x} = p_{x}/m Solution:- To obtain the area, we have to use Simpson’s rule. Using Simpson’s rule, the area will be, J_{x} = 1/3 h (f_{0} +4f_{1}+2f_{2}+4f_{3}+……+4f_{1}3 + f_{1}4) Here h (=0.2 ms) is the width of each strip. So, J_{x} = 1/3 h (f_{0} +4f_{1}+2f_{2}+4f_{3}+……+4f_{1}3 + f_{1}4) = 1/3 (0.2 ms) (200+4(800)+2(1200))N = 1/3 (0.2 ms×10^{-3} s/1 ms) (200+4(800)+2(1200))N = (4.28 N.s) (1 kg.m/s^{2}/ 1 N) = 4.28 kg.m/s The impulse is the change in momentum and the ball started from rest. So the initial momentum of the ball will be zero (p_{ix}=0). To obtain the final momentum p_{fx}, substitute 4.28 kg.m/s for J_{x} and 0 m/s for p_{ix} in the equation p_{fx} = J_{x} + p_{ix}, p_{fx} = J_{x} + p_{ix} = 4.28 kg.m/s + 0 m/s = 4.28 kg.m/s To obtain the final velocity v_{fx}, substitute 4.28 kg.m/s for p_{fx }and 0.50 kg for m in the equation v_{fx} = p_{fx}/m , v_{fx} = p_{fx} = (4.28 kg.m/s) / (0.5 kg) = 8.6 m/s From the above observation we conclude that, the final velocity v_{fx }of the ball will be 8.6 m/s. ________________________________________________________________ Problem 4:- Meteor Crater in Arizona as shown in the below figure is thought to have been formed by the impact of a meteorite with the Earth some 20,000 years ago. The mass of the meteorite is estimated to be 5×10^{10} kg and its speed to have been 7.2 km/s. What speed would such a meteorite impart to the Earth in a head-on collision? Concept:- Momentum of the body p is equal to the mass of the body m times velocity of the body v. So, p = mv In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system. Consider the initial momentum of the Earth is p_{i,1}, initial momentum of the meteorite is p_{i,2}, final momentum of the Earth is p_{f,1} and final momentum of the meteorite is p_{f,2}. To obtain the initial momentum p_{i,1} of the Earth, substitute p_{i,1} for p, m_{1} for the mass of the Earth and v_{1,i} for the initial velocity of the Earth in the equation p = mv, p_{i,1} = m_{1} v_{1,i} To obtain the initial momentum p_{i,2} of the meteorite, substitute p_{i,2} for p, m_{2} for the mass of the meteorite and v_{2,i} for the initial velocity of the meteorite in the equation p = mv, p_{i,2} = m_{2} v_{2,i} The total initial momentum of the meteorite and the Earth is p_{i,1 }+ p_{i,2} = m_{1} v_{1,i} + m_{2} v_{2,i} It is given that the collision is inelastic. This means the system of mass of the meteorite and the earth moves with a common velocity. This velocity is equal to the velocity of the meteorite before the impact. So, the final momentum of the system will be, p_{f }= (m_{1} + m_{2}) v_{f} So applying conservation of momentum to this system, the sum of the initial momentum of the Earth and meteorite will be equal to the sum of the final momentum of the Earth and meteorite. p_{i,1 }+ p_{i,2} = p_{f} m_{1} v_{1,i} + m_{2} v_{2,i} = (m_{1} + m_{2}) v_{f} v_{f} = (m_{1} v_{1,i} + m_{2} v_{2,i}) / (m_{1} + m_{2}) Solution:- Let us consider the velocity of the earth with respect to the meteorite is zero. Thus, v_{1,i }= 0 So, v_{f }= (0+ m_{2} v_{2,i}) / (m_{1} + m_{2}) = (m_{2} v_{2,i}) / (m_{1} + m_{2}) To obtain the speed v_{f} of the meteorite impart to the earth in a head-on collision, substitute 5×10^{10} kg for m_{2}, 5.98×10^{24} kg for m_{1} and 7200 m/s for v_{2,I} in the equation v_{f }= (m_{2} v_{2,i}) / (m_{1} + m_{2}), v_{f }= (m_{2} v_{2,i}) / (m_{1} + m_{2}) = (5×10^{10} kg) (7200 m/s) /(5×10^{10} kg)+ (5.98×10^{24} kg) = 7×10^{-11} m/s From the above observation we conclude that, the speed vf of the meteorite impart to the earth in a head-on collision would be 7×10-11 m/s. _______________________________________________________________ Problem 5:- Spacecraft Voyager 2 (mass m and speed v relative to the Sun) approaches the planet Jupiter (mass M and speed V relative to the sun) as shown in the below figure. The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this “slingshot” encounter? Assume that v = 12 km/s and V = 13 km/s (the orbital speed of Jupiter), and that this is an elastic collision. The mass of Jupiter is very much greater than the mass of the spacecraft, M >>m. Concept:- In accordance to the law of conservation of momentum, the total momentum of the particle having mass m_{1} and the particle having mass m_{2} before the collision equals their total momentum after the collision. The changes in momentum of the two objects have equal magnitudes and opposite signs, a necessary consequence of the law of conservation of momentum. In an elastic collision, the final velocity v_{1f} of the body having mass m_{1} is, v_{1f} = (m_{1} – m_{2}/ m_{1} + m_{2}) v_{1i} + (2m_{2}/ m_{1} + m_{2}) v_{2i} and the final velocity v_{2f} of the body having mass m_{2} is, v_{2f} = (2m_{1}/ m_{1} + m_{2}) v_{1i} + (m_{2} – m_{1}/ m_{1} + m_{2}) v_{2i} Here, v_{1i} is the initial velocity of the body having mass m_{1} and v_{2i} is the initial velocity of the body having mass m_{2}. When, m_{1 >> }m_{2}, then, v_{1f} = v_{1i} and v_{2f} ≈ 2v_{1i} - v_{2i} Solution:- Since the mass of Jupiter M (m_{1} = M) is very much greater than the mass of the spacecraft m (m_{2} = m), therefore the speed of the spacecraft v_{2f} relative to the Sun will be, v_{2f} ≈ 2v_{1i} - v_{2i} ≈ 2V – v (Since, V = v_{1i} and v =v_{2i}) Here, speed of the Jupiter relative to the Sun is V (=v_{1i}) and speed of the spacecraft relative to the Sun is v (=v_{2i}). To obtain the speed v_{2f} of the spacecraft relative to the Sun, after the “slingshot” encounter, substitute 13 km/s for speed of the Jupiter relative to the Sun V and -12 km/s for speed of the spacecraft relative to the Sun v in the equation v_{2f} = 2V – v, v_{2f} = 2V – v = 2(13 km/s) – (-12 m/s) = (26 km/s) + (12 km/s) = 38 km/s From the above observation we conclude that, the speed v_{2f} of the spacecraft relative to the Sun, after the “slingshot” encounter would be 38 km/s. _______________________________________________________ Probelm 6:- It is well known that bullets and other missiles fired at Super man simply bounce off his chest as shown in below figure. Suppose that a gangster sprays Superman’s chest with 3.0-g bullets at the rate of 100 bullets/min, the speed of each bullet being 500 m/s. Suppose too that the bullets rebound straight back with no loss in speed. Find the average force exerted by the stream of bullets on Superman’s chest. Concept:- The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval. So, J = Δp = p_{f} - p_{i} Here p_{f} is the final momentum and p_{i} is the initial momentum of the body. Impulse of a force J is defined as, J = F_{av} t Here F_{av} is the average force and t is the impact time. J = F_{av} t So, F_{av} = J/t Solution:- The change in momentum Δp of one bullet will be, Δp = p_{f} - p_{i} = mv- (-mv) = 2mv To obtain the change in momentum Δp of one bullet, substitute 3.0 g for mass m and 500 m/s for v in the equation Δp = 2mv, Δp = 2mv = 2(3.0 g) (500 m/s) = 2(3.0 g×10^{-3} kg/1 g) (500 m/s) = 3.0 kg.m/s The average force F_{av} is equal to the total impulse J in one minute divided by one minute. So, F_{av} = 100(J)/t = 100(Δp)/t To find out the average force F_{av} exerted by the stream of bullets on Superman’s chest, substitute 3.0 kg.m/s for Δp and 60 s for t in the equation F_{av} = 100(Δp)/t, F_{av} =100(Δp)/t =100(3.0 kg.m/s)/(60 s) =(5.0 kg.m/s^{2}) (1 N/1 kg.m/s^{2}) = 5.0 N Thus from the above observation we conclude that, the average force F_{av} exerted by the stream of bullets on Superman’s chest would be 5.0 N. ______________________________________________________________ Problem 7:- The two spheres on the right of below figure are slightly separated and initially at rest; the left sphere is incident with speed v_{0}. Assuming head-on elastic collision, (a) if m ≥ M, show that there are two collisions and find all final velocities; (b) if m<M, show that there are three collisions and all final velocities. Concept:- In accordance to the law of conservation of momentum, the total momentum of the particle having mass m_{1} and the particle having mass m_{2} before the collision equals their total momentum after the collision. The changes in momentum of the two objects have equal magnitudes and opposite signs, a necessary consequence of the law of conservation of momentum. In an elastic collision, the final velocity v_{1f} of the body having mass m_{1} is, v_{1f} = (m_{1} – m_{2}/ m_{1} + m_{2}) v_{1i} + (2m_{2}/ m_{1} + m_{2}) v_{2i} and the final velocity v_{2f} of the body having mass m_{2} is, v_{2f} = (2m_{1}/ m_{1} + m_{2}) v_{1i} + (m_{2} – m_{1}/ m_{1} + m_{2}) v_{2i} Here, v_{1i} is the initial velocity of the body having mass m_{1} and v_{2i} is the initial velocity of the body having mass m_{2}. Solution:- As the target particle second cart (m_{2}) is at rest, thus, v_{2i} = 0. Thus, v_{1f} = (m_{1} – m_{2}/ m_{1} + m_{2}) v_{1i} and v_{2f} = (2m_{1}/ m_{1} + m_{2}) v_{1i} The two spheres in the below figure are slightly separated and initially at rest; the left sphere is incident with speed v_{0}. As head-on collision occurs so there will always be at least two collisions. Consider the balls are a,b, and c from left to right. After the first collision between a and b one has, the velocity of b will be, v_{b}_{,1}= v_{0} and the velocity of a will be, v_{a}_{,1}= 0 After the first collision between b and c, the velocity of c will be, v_{c}_{,1}=2mv_{0}/m+M and the velocity of b will be, v_{b}_{,2}=(m-M)v_{0}/(m+M) (a) If m ≥ M then the ball b continue to move to right (or stops) and there are no more collisions. (b) If m < M then ball b bounces back and strikes ball a which was at rest. Then, v_{a}_{,2} = (m-M)v_{0}/(m+M) and v_{b}_{,3}= 0 __________________________________________________________________ Problem 8:- Two cars A and B slide on an icy road as they attempt to stop at a traffic light. The mass of A is 1100 kg and the mass of B is 1400 kg. The coefficient of kinetic friction between the locked wheels of both cars and the road is 0.130. Car A succeeds in coming to rest at the light, but car B cannot stop and rear-ends car A. After the collision, A comes to rest 8.20 m ahead of the impact point and B 6.10 m ahead: as shown in the below figure. Both drivers had their brakes locked throughout the incident. (a) From the distances each car moved after the collision, find the speed of each car immediately after impact. (b) Use conservation of momentum to find the speed at which car B struck car A. On what grounds can the use of momentum conservation be criticized here? Concept:- Acceleration a due to friction is defined as, a =µ_{k}g Here µ_{k} is the coefficient of kinetic friction and g is the free fall acceleration. Time t is equal to the speed v of the object divided by the acceleration a of the object. t = v/a In accordance to equation of motion, the distance d traveled by the object is equal to, d = ut+1/2at^{2} Here u is the initial velocity, t is the time and a is the acceleration. As the initial velocity u=0, thus the equation d = ut+1/2at^{2} will become, d = ut+1/2at^{2} = (0)t+1/2at^{2} = 1/2at^{2} Solution:- (a) For an object with initial speed v and deceleration –a which travels a distance x before stopping. So the time t to stop will be, t=v/a The average speed while stopping is v/2. The distance x will be, x=1/2at^{2} To obtain the speed v, substitute v/a for t in the equation d=1/2at^{2}, we get, x=1/2at^{2} = 1/2a(v/a)^{2} 2x=v^{2}/a Or,v=√2ax To find out speed in terms of coefficient friction, Substitute µ_{k}g for a in the equation v=√2ax, v=√2ax =√2 (µ_{k}g) x =√2µ_{k}g x To obtain the speed of car A after the collision, substitute 0.130 for µ_{k}, 9.81m/s^{2} for g and 8.20 m for x in the equation v=√2µ_{k}g x, we get, v=√2µ_{k}g x =√2(0.130)(9.81m/s^{2})(8.20 m) =4.57 m/s To obtain the speed of car B after the collision, substitute 0.130 for µ_{k}, 9.81 m/s^{2} for g and 6.10 m for x in the equation v=√2µ_{k}g x, we get, v=√2µ_{k}g x =√2(0.130)(9.81m/s^{2})(6.10 m) =3.94 m/s From the above observation we conclude that, the speed of car A after the collision will be 4.57 m/s while the speed of car B will be 3.94 m/s. (b) If v_{0} is the at which car B struck car A, then in accordance to law of conservation of linear momentum, m_{B} v_{0} = m_{A}v_{A}+ m_{B}v_{B} So, v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B} Here m_{A} is the mass of car A, m_{B} is the mass of car B, v_{A} is the speed of car A and v_{B} is the speed of car B. To obtain the speed v_{0} at which car B struck car A, substitute 1100 kg for m_{A}, 4.57 m/s for v_{A}, 1400 kg for m_{B} and 3.94 m/s for v_{B} in the equation v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B}, v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B} =[(1100 kg)(4.57 m/s)+(1400 kg)(3.94 m/s)]/(1400 kg) =7.53 m/s From the above observation we conclude that, the speed v_{0} at which car B struck car A would be 7.53 m/s. _________________________________________________________________ Problem 9:- A karate expert breakes a pine board, 2.2 cm thick, with a hand chop. Strobe photography shows that the hand, whose mass may be taken as 540 g, strikes the top of the board with a speed of 9.5 m/s and comes to rest 2.8 cm below this level. (a) What is the time duration of the shop (assuming a constant force)? (b) What average force is applied? Concept:- Average speed v_{av} is equal to the total distance y travelled by the body divided by total time t taken by the body to travel that distance. So, v_{av} = y/t Thus the time t will be, t = y/ v_{av} Momentum of a body (p) is defined as, p = mv Here, m is the mass of the body and v is the velocity of the body. The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval. So, J = Δp = mΔv (Since, Δp = mΔv) Impulse of a force J is defined as, J = F_{av} t Here F_{av} is the average force and t is the impact time. J = F_{av} t So, F_{av} = J/t Solution:- (a) The average speed v_{av} during the time the hand is in contact with the board is half of the initial speed. Thus, v_{av} = (9.5 m/s) /2 = 4.8 m/s To obtain the time duration t of the chop, substitute 2.8 cm for y and 4.8 m/s for v_{av} in the equation t = y/ v_{av}, t = y/ v_{av} = (2.8 cm)/(4.8 m/s) = (2.8 cm×10^{-2} m/1 cm)/(4.8 m/s) = 5.8×10^{-3} s = (5.8×10^{-3} s) (1 ms/10^{-3} s) = 5.8 ms Therefore the time duration t of the chop will be 5.8 ms. (b) To find out the applied average force F_{av}, first we have to find out the impulse of force J. The impulse given to the board is the same as the magnitude in the change in momentum of the hand. To obtain impulse J given to the board, substitute 540 g for m and 9.5 m/s for Δv in the equation J = mΔv, J = mΔv = (540 g) (9.5 m/s) = (540 g× 1 kg/10^{3}g) (9.5 m/s) = (5.1 kg.m/s) (1 N/1 kg.m/s^{2}) = 5.1 N.s To obtain the applied average force F_{av}, substitute 5.1 N.s for J and 5.8 ms for t in the equation F_{av} = J/t, F_{av} = J/t =(5.1 N.s) / (5.8 ms) = (5.1 N.s) / (5.8 ms×10^{-3} s/1 ms) = 880 N From the above observation we conclude that, the applied average force F_{av} would be 830 N. __________________________________________________________________________ Problem 10:- A 195-lb man standing on a surface of negligible friction kicks forward a 0.158-lb stone lying at his feet so that it acquires a speed of 12.7 ft/s. What velocity does the man acquires as a result? Concept:- Law of conservation of linear momentum states that, in an isolated system (no external force), the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other. Therefore total momentum of the system before collision is equal to the total momentum of the system after collision. The momentum of particle p is equal to the mass of particle m times the velocity of particle v. So, p = mv …… (1) Solution:- Since both the body is initially rest, therefore the velocities of both the bodies are zero resulting the initial momentum of the system will be zero. So, initial momentum of the system = 0 Let us consider m_{m} is the mass of the man and v_{m} is the velocity of the man when the man kicks the stone forward. So using equation (1), the momentum of the man (p_{m}) when he kicks the stone forward will be, p_{m}= m_{m}v_{m} …… (2) Let us consider m_{s} is the mass of the stone and v_{s} is the velocity of the stone after kicked by the man. So again using equation (1), the momentum of the stone (p_{2}) after kicked by the man will be, p_{s}= m_{s}v_{s} …… (3) Final momentum of the system = p_{m} + p_{s} = m_{m}v_{m} + m_{s}v_{s} …… (4) Conservation of linear momentum states that, the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other. Therefore total momentum of the system before collision is equal to the total momentum of the system after collision. So, p_{m} + p_{s} = 0 So, m_{m}v_{m} + m_{s}v_{s} = 0 m_{m}v_{m} = -m_{s}v_{s} v_{m} = -m_{s}v_{s}/m_{m} …… (5) To obtain the velocity of the man, substitute 0.158-lb for mass of the stone m_{s}, 12.7 ft/s for speed of the stone v_{s} and 195-lb for mass of the man m_{m} in the equation v_{m} = -m_{s}v_{s}/m_{m}. So, v_{m} = -m_{s}v_{s}/m_{m} = - (0.158-lb) (12.7 ft/s)/(195-lb) = -1.0×10^{-2} ft/s …… (6) From equation (5) we observed that, the velocity of the man will be -1.0×10^{-2} ft/s. Related Resources:- You might like to refer Conservation of Momentum. For getting an idea of the type of questions asked, refer the Previous Year Question Papers. Click here to refer the most Useful Books of Physics.
A 4.88-kg object with a speed of 3.14 m/s strikes a steel plate at an angle of 42.0º and rebounds at the same speed and angle as shown in the below figure. What is the change (magnitude and direction) of the linear momentum of the object?
Momentum of the object (p) is defined as the mass of the object (m) time’s velocity of the object (v).
p = mv …… (1)
The below figure shows a 4.88-kg object with a speed of 31.4 m/s strikes a steel plate at an angle of 42.0^{°} and rebounds at the same speed with same angle which is shown geometrically.
The initial momentum of the object is p_{i} when it strikes the steel plate.
To find initial momentum p_{i}, substitute p_{i} for p, 4.88-kg for mass m of the object and 31.4 m/s for velocity of the object in the equation p = mv,
p_{i} = mv
= (4.88 kg) (31.4 m/s)
= 153 kg. m/s …… (2)
Since the final momentum p_{f} of the object is equal to the initial momentum p_{i} of the object, therefore,
p_{f} = 153 kg. m/s …… (3)
In the above figure the angle θ will be,
θ = 42° + 42°
= 84°
Using cosine law, the magnitude of change of the linear momentum of the object Δp will be,
Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f})cos θ …… (4)
To obtain magnitude of linear momentum Δp, substitute 153 kg. m/s for p_{i}, 153 kg. m/s for p_{f} and 84° for the angle θ in the equation Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f}) cos θ,
Δp = √( p_{i})^{2} + (p_{f})^{2} + 2 (-p_{i})(p_{f}) cos θ
= √(153 kg. m/s)^{2} + (153 kg. m/s)^{2} - 2 (153 kg. m/s)( 153 kg. m/s) cos 84°
= 205 kg. m/s
From the above observation we conclude that, the change of the linear momentum of the object will be, 205 kg. m/s and also from the figure the direction of change of the linear momentum of the object is perpendicular to the steel plate.
____________________________________________________________
Below figure shows an approximate representation of force versus time during the collision of a 58-g tennis ball with a wall. The initial velocity of the ball is 32 m/s perpendicular to the wall; it rebounds with the same speed, also perpenducular to the wall. What is the value of F_{max}, the maximum contact force during the collision?
Momentum of a body (p) is defined as,
p = mv
Here, m is the mass of the body and v is the velocity of the body.
The impulse of the net force J acting on a particle during a given time interval is equal to the change in momentum Δp of the particle during that interval.
So, J = Δp
= p_{f} - p_{i}
Here p_{f} is the final momentum and p_{i} is the initial momentum of the body.
Again the impulse is the area under a force-time graph.
As the tennis ball strikes with the wall having velocity v and then rebounds with same velocity, thus the change in momentum Δp of the tennis ball will be equal to,
Δp = (mv) – (-mv)
= 2mv
To obtain the change in momentum Δp of the ball, substitute 58-g for m and 32 m/s for v in the equation Δp = 2mv, we get,
Δp = 2mv
= 2 (58-g) (32 m/s)
= 2 (58-g×10^{-3} kg/1 g) (32 m/s)
= 3.7 kg.m/s
To obtain the impulse we have to find out the area under force-time graph for the trapezoid.
So, J = F_{max} (2 ms+6 ms)/2
= F_{max} (4 ms)
So, F_{max} = J/4 ms
= Δp/4 ms
To obtain the maximum contact force F_{max} during the collision, substitute 3.7 kg.m/s for Δp in the equation F_{max}= Δp/4 ms,
F_{max}= Δp/4 ms
= (3.7 kg.m/s) / (4 ms)
= (3.7 kg.m/s) / (4 ms×10^{-3} s/1 ms)
= 930 kg. m/s^{2}
= (930 kg. m/s^{2}) (1 N/1 kg. m/s^{2})
= 930 N
From the above observation we conclude that, the maximum contact force F_{max} during the collision would be 930 N.
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A croquet ball with a mass 0.50 kg is struck by a mallet, receiving the impulse shown in the graph below. What is the ball’s velocity just after the force has become zero?
J_{x} = Δp
= p_{fx} – p_{ix}
Here p_{fx} is the final momentum and p_{ix} is the initial momentum of the body.
So, p_{fx} = J_{x} + p_{ix}
Momentum p_{x} of a particle is equal to the mass of the particle times velocity v_{x} of the particle.
So, p_{x} = m v_{x}
Or, v_{x} = p_{x}/m
To obtain the area, we have to use Simpson’s rule.
Using Simpson’s rule, the area will be,
J_{x} = 1/3 h (f_{0} +4f_{1}+2f_{2}+4f_{3}+……+4f_{1}3 + f_{1}4)
Here h (=0.2 ms) is the width of each strip.
So, J_{x} = 1/3 h (f_{0} +4f_{1}+2f_{2}+4f_{3}+……+4f_{1}3 + f_{1}4)
= 1/3 (0.2 ms) (200+4(800)+2(1200))N
= 1/3 (0.2 ms×10^{-3} s/1 ms) (200+4(800)+2(1200))N
= (4.28 N.s) (1 kg.m/s^{2}/ 1 N)
= 4.28 kg.m/s
The impulse is the change in momentum and the ball started from rest. So the initial momentum of the ball will be zero (p_{ix}=0).
To obtain the final momentum p_{fx}, substitute 4.28 kg.m/s for J_{x} and 0 m/s for p_{ix} in the equation p_{fx} = J_{x} + p_{ix},
p_{fx} = J_{x} + p_{ix}
= 4.28 kg.m/s + 0 m/s = 4.28 kg.m/s
To obtain the final velocity v_{fx}, substitute 4.28 kg.m/s for p_{fx }and 0.50 kg for m in the equation v_{fx} = p_{fx}/m ,
v_{fx} = p_{fx}
= (4.28 kg.m/s) / (0.5 kg)
= 8.6 m/s
From the above observation we conclude that, the final velocity v_{fx }of the ball will be 8.6 m/s.
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Meteor Crater in Arizona as shown in the below figure is thought to have been formed by the impact of a meteorite with the Earth some 20,000 years ago. The mass of the meteorite is estimated to be 5×10^{10} kg and its speed to have been 7.2 km/s. What speed would such a meteorite impart to the Earth in a head-on collision?
Momentum of the body p is equal to the mass of the body m times velocity of the body v.
So, p = mv
In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system.
Consider the initial momentum of the Earth is p_{i,1}, initial momentum of the meteorite is p_{i,2}, final momentum of the Earth is p_{f,1} and final momentum of the meteorite is p_{f,2}.
To obtain the initial momentum p_{i,1} of the Earth, substitute p_{i,1} for p, m_{1} for the mass of the Earth and v_{1,i} for the initial velocity of the Earth in the equation p = mv,
p_{i,1} = m_{1} v_{1,i}
To obtain the initial momentum p_{i,2} of the meteorite, substitute p_{i,2} for p, m_{2} for the mass of the meteorite and v_{2,i} for the initial velocity of the meteorite in the equation p = mv,
p_{i,2} = m_{2} v_{2,i}
The total initial momentum of the meteorite and the Earth is
p_{i,1 }+ p_{i,2} = m_{1} v_{1,i} + m_{2} v_{2,i}
It is given that the collision is inelastic. This means the system of mass of the meteorite and the earth moves with a common velocity. This velocity is equal to the velocity of the meteorite before the impact.
So, the final momentum of the system will be,
p_{f }= (m_{1} + m_{2}) v_{f}
So applying conservation of momentum to this system, the sum of the initial momentum of the Earth and meteorite will be equal to the sum of the final momentum of the Earth and meteorite.
p_{i,1 }+ p_{i,2} = p_{f}
m_{1} v_{1,i} + m_{2} v_{2,i} = (m_{1} + m_{2}) v_{f}
v_{f} = (m_{1} v_{1,i} + m_{2} v_{2,i}) / (m_{1} + m_{2})
Let us consider the velocity of the earth with respect to the meteorite is zero.
Thus,
v_{1,i }= 0
So, v_{f }= (0+ m_{2} v_{2,i}) / (m_{1} + m_{2})
= (m_{2} v_{2,i}) / (m_{1} + m_{2})
To obtain the speed v_{f} of the meteorite impart to the earth in a head-on collision, substitute 5×10^{10} kg for m_{2}, 5.98×10^{24} kg for m_{1} and 7200 m/s for v_{2,I} in the equation v_{f }= (m_{2} v_{2,i}) / (m_{1} + m_{2}),
v_{f }= (m_{2} v_{2,i}) / (m_{1} + m_{2})
= (5×10^{10} kg) (7200 m/s) /(5×10^{10} kg)+ (5.98×10^{24} kg)
= 7×10^{-11} m/s
From the above observation we conclude that, the speed vf of the meteorite impart to the earth in a head-on collision would be 7×10-11 m/s.
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Spacecraft Voyager 2 (mass m and speed v relative to the Sun) approaches the planet Jupiter (mass M and speed V relative to the sun) as shown in the below figure. The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this “slingshot” encounter? Assume that v = 12 km/s and V = 13 km/s (the orbital speed of Jupiter), and that this is an elastic collision. The mass of Jupiter is very much greater than the mass of the spacecraft, M >>m.
In accordance to the law of conservation of momentum, the total momentum of the particle having mass m_{1} and the particle having mass m_{2} before the collision equals their total momentum after the collision. The changes in momentum of the two objects have equal magnitudes and opposite signs, a necessary consequence of the law of conservation of momentum.
In an elastic collision, the final velocity v_{1f} of the body having mass m_{1} is,
v_{1f} = (m_{1} – m_{2}/ m_{1} + m_{2}) v_{1i} + (2m_{2}/ m_{1} + m_{2}) v_{2i}
and the final velocity v_{2f} of the body having mass m_{2} is,
v_{2f} = (2m_{1}/ m_{1} + m_{2}) v_{1i} + (m_{2} – m_{1}/ m_{1} + m_{2}) v_{2i}
Here, v_{1i} is the initial velocity of the body having mass m_{1} and v_{2i} is the initial velocity of the body having mass m_{2}.
When, m_{1 >> }m_{2}, then,
v_{1f} = v_{1i} and v_{2f} ≈ 2v_{1i} - v_{2i}
Since the mass of Jupiter M (m_{1} = M) is very much greater than the mass of the spacecraft m (m_{2} = m), therefore the speed of the spacecraft v_{2f} relative to the Sun will be,
v_{2f} ≈ 2v_{1i} - v_{2i}
≈ 2V – v (Since, V = v_{1i} and v =v_{2i})
Here, speed of the Jupiter relative to the Sun is V (=v_{1i}) and speed of the spacecraft relative to the Sun is v (=v_{2i}).
To obtain the speed v_{2f} of the spacecraft relative to the Sun, after the “slingshot” encounter, substitute 13 km/s for speed of the Jupiter relative to the Sun V and -12 km/s for speed of the spacecraft relative to the Sun v in the equation v_{2f} = 2V – v,
v_{2f} = 2V – v
= 2(13 km/s) – (-12 m/s)
= (26 km/s) + (12 km/s)
= 38 km/s
From the above observation we conclude that, the speed v_{2f} of the spacecraft relative to the Sun, after the “slingshot” encounter would be 38 km/s.
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It is well known that bullets and other missiles fired at Super man simply bounce off his chest as shown in below figure. Suppose that a gangster sprays Superman’s chest with 3.0-g bullets at the rate of 100 bullets/min, the speed of each bullet being 500 m/s. Suppose too that the bullets rebound straight back with no loss in speed. Find the average force exerted by the stream of bullets on Superman’s chest.
Impulse of a force J is defined as,
J = F_{av} t
Here F_{av} is the average force and t is the impact time.
So, F_{av} = J/t
The change in momentum Δp of one bullet will be,
Δp = p_{f} - p_{i}
= mv- (-mv)
To obtain the change in momentum Δp of one bullet, substitute 3.0 g for mass m and 500 m/s for v in the equation Δp = 2mv,
= 2(3.0 g) (500 m/s)
= 2(3.0 g×10^{-3} kg/1 g) (500 m/s)
= 3.0 kg.m/s
The average force F_{av} is equal to the total impulse J in one minute divided by one minute.
So, F_{av} = 100(J)/t
= 100(Δp)/t
To find out the average force F_{av} exerted by the stream of bullets on Superman’s chest, substitute 3.0 kg.m/s for Δp and 60 s for t in the equation F_{av} = 100(Δp)/t,
F_{av} =100(Δp)/t
=100(3.0 kg.m/s)/(60 s)
=(5.0 kg.m/s^{2}) (1 N/1 kg.m/s^{2})
= 5.0 N
Thus from the above observation we conclude that, the average force F_{av} exerted by the stream of bullets on Superman’s chest would be 5.0 N.
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The two spheres on the right of below figure are slightly separated and initially at rest; the left sphere is incident with speed v_{0}. Assuming head-on elastic collision, (a) if m ≥ M, show that there are two collisions and find all final velocities; (b) if m<M, show that there are three collisions and all final velocities.
As the target particle second cart (m_{2}) is at rest, thus, v_{2i} = 0.
Thus, v_{1f} = (m_{1} – m_{2}/ m_{1} + m_{2}) v_{1i} and v_{2f} = (2m_{1}/ m_{1} + m_{2}) v_{1i}
The two spheres in the below figure are slightly separated and initially at rest; the left sphere is incident with speed v_{0}. As head-on collision occurs so there will always be at least two collisions. Consider the balls are a,b, and c from left to right.
After the first collision between a and b one has, the velocity of b will be,
v_{b}_{,1}= v_{0}
and
the velocity of a will be,
v_{a}_{,1}= 0
After the first collision between b and c, the velocity of c will be,
v_{c}_{,1}=2mv_{0}/m+M
the velocity of b will be,
v_{b}_{,2}=(m-M)v_{0}/(m+M)
(a) If m ≥ M then the ball b continue to move to right (or stops) and there are no more collisions.
(b) If m < M then ball b bounces back and strikes ball a which was at rest.
Then,
v_{a}_{,2} = (m-M)v_{0}/(m+M)
v_{b}_{,3}= 0
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Two cars A and B slide on an icy road as they attempt to stop at a traffic light. The mass of A is 1100 kg and the mass of B is 1400 kg. The coefficient of kinetic friction between the locked wheels of both cars and the road is 0.130. Car A succeeds in coming to rest at the light, but car B cannot stop and rear-ends car A. After the collision, A comes to rest 8.20 m ahead of the impact point and B 6.10 m ahead: as shown in the below figure. Both drivers had their brakes locked throughout the incident. (a) From the distances each car moved after the collision, find the speed of each car immediately after impact. (b) Use conservation of momentum to find the speed at which car B struck car A. On what grounds can the use of momentum conservation be criticized here?
Acceleration a due to friction is defined as,
a =µ_{k}g
Here µ_{k} is the coefficient of kinetic friction and g is the free fall acceleration.
Time t is equal to the speed v of the object divided by the acceleration a of the object.
t = v/a
In accordance to equation of motion, the distance d traveled by the object is equal to,
d = ut+1/2at^{2}
Here u is the initial velocity, t is the time and a is the acceleration.
As the initial velocity u=0, thus the equation d = ut+1/2at^{2} will become,
= (0)t+1/2at^{2}
= 1/2at^{2}
(a) For an object with initial speed v and deceleration –a which travels a distance x before stopping.
So the time t to stop will be,
t=v/a
The average speed while stopping is v/2.
The distance x will be,
x=1/2at^{2}
To obtain the speed v, substitute v/a for t in the equation d=1/2at^{2}, we get,
= 1/2a(v/a)^{2}
2x=v^{2}/a
Or,v=√2ax
To find out speed in terms of coefficient friction, Substitute µ_{k}g for a in the equation v=√2ax,
v=√2ax
=√2 (µ_{k}g) x
=√2µ_{k}g x
To obtain the speed of car A after the collision, substitute 0.130 for µ_{k}, 9.81m/s^{2} for g and 8.20 m for x in the equation v=√2µ_{k}g x, we get,
v=√2µ_{k}g x
=√2(0.130)(9.81m/s^{2})(8.20 m)
=4.57 m/s
To obtain the speed of car B after the collision, substitute 0.130 for µ_{k}, 9.81 m/s^{2} for g and 6.10 m for x in the equation v=√2µ_{k}g x, we get,
=√2(0.130)(9.81m/s^{2})(6.10 m)
=3.94 m/s
From the above observation we conclude that, the speed of car A after the collision will be 4.57 m/s while the speed of car B will be 3.94 m/s.
(b) If v_{0} is the at which car B struck car A, then in accordance to law of conservation of linear momentum,
m_{B} v_{0} = m_{A}v_{A}+ m_{B}v_{B}
So, v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B}
Here m_{A} is the mass of car A, m_{B} is the mass of car B, v_{A} is the speed of car A and v_{B} is the speed of car B.
To obtain the speed v_{0} at which car B struck car A, substitute 1100 kg for m_{A}, 4.57 m/s for v_{A}, 1400 kg for m_{B} and 3.94 m/s for v_{B} in the equation v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B},
v_{0} = m_{A}v_{A}+ m_{B}v_{B}/ m_{B}
=[(1100 kg)(4.57 m/s)+(1400 kg)(3.94 m/s)]/(1400 kg)
=7.53 m/s
From the above observation we conclude that, the speed v_{0} at which car B struck car A would be 7.53 m/s.
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A karate expert breakes a pine board, 2.2 cm thick, with a hand chop. Strobe photography shows that the hand, whose mass may be taken as 540 g, strikes the top of the board with a speed of 9.5 m/s and comes to rest 2.8 cm below this level. (a) What is the time duration of the shop (assuming a constant force)? (b) What average force is applied?
Average speed v_{av} is equal to the total distance y travelled by the body divided by total time t taken by the body to travel that distance.
So, v_{av} = y/t
Thus the time t will be,
t = y/ v_{av}
= mΔv (Since, Δp = mΔv)
(a) The average speed v_{av} during the time the hand is in contact with the board is half of the initial speed.
Thus, v_{av} = (9.5 m/s) /2 = 4.8 m/s
To obtain the time duration t of the chop, substitute 2.8 cm for y and 4.8 m/s for v_{av} in the equation t = y/ v_{av},
= (2.8 cm)/(4.8 m/s)
= (2.8 cm×10^{-2} m/1 cm)/(4.8 m/s)
= 5.8×10^{-3} s
= (5.8×10^{-3} s) (1 ms/10^{-3} s)
= 5.8 ms
Therefore the time duration t of the chop will be 5.8 ms.
The impulse given to the board is the same as the magnitude in the change in momentum of the hand.
To obtain impulse J given to the board, substitute 540 g for m and 9.5 m/s for Δv in the equation J = mΔv,
J = mΔv
= (540 g) (9.5 m/s)
= (540 g× 1 kg/10^{3}g) (9.5 m/s)
= (5.1 kg.m/s) (1 N/1 kg.m/s^{2})
= 5.1 N.s
To obtain the applied average force F_{av}, substitute 5.1 N.s for J and 5.8 ms for t in the equation F_{av} = J/t,
F_{av} = J/t
=(5.1 N.s) / (5.8 ms)
= (5.1 N.s) / (5.8 ms×10^{-3} s/1 ms)
= 880 N
From the above observation we conclude that, the applied average force F_{av} would be 830 N.
__________________________________________________________________________
A 195-lb man standing on a surface of negligible friction kicks forward a 0.158-lb stone lying at his feet so that it acquires a speed of 12.7 ft/s. What velocity does the man acquires as a result?
Law of conservation of linear momentum states that, in an isolated system (no external force), the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other. Therefore total momentum of the system before collision is equal to the total momentum of the system after collision.
The momentum of particle p is equal to the mass of particle m times the velocity of particle v.
So, p = mv …… (1)
Since both the body is initially rest, therefore the velocities of both the bodies are zero resulting the initial momentum of the system will be zero.
So, initial momentum of the system = 0
Let us consider m_{m} is the mass of the man and v_{m} is the velocity of the man when the man kicks the stone forward.
So using equation (1), the momentum of the man (p_{m}) when he kicks the stone forward will be,
p_{m}= m_{m}v_{m} …… (2)
Let us consider m_{s} is the mass of the stone and v_{s} is the velocity of the stone after kicked by the man.
So again using equation (1), the momentum of the stone (p_{2}) after kicked by the man
will be,
p_{s}= m_{s}v_{s} …… (3)
Final momentum of the system = p_{m} + p_{s}
= m_{m}v_{m} + m_{s}v_{s} …… (4)
Conservation of linear momentum states that, the algebraic sum of momenta of bodies, along any straight line, remains constant and is not changed due to their mutual action and reaction on each other. Therefore total momentum of the system before collision is equal to the total momentum of the system after collision.
So,
p_{m} + p_{s} = 0
So, m_{m}v_{m} + m_{s}v_{s} = 0
m_{m}v_{m} = -m_{s}v_{s}
v_{m} = -m_{s}v_{s}/m_{m} …… (5)
To obtain the velocity of the man, substitute 0.158-lb for mass of the stone m_{s}, 12.7 ft/s for speed of the stone v_{s} and 195-lb for mass of the man m_{m} in the equation v_{m} = -m_{s}v_{s}/m_{m}.
So, v_{m} = -m_{s}v_{s}/m_{m}
= - (0.158-lb) (12.7 ft/s)/(195-lb)
= -1.0×10^{-2} ft/s …… (6)
From equation (5) we observed that, the velocity of the man will be -1.0×10^{-2} ft/s.
You might like to refer Conservation of Momentum.
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