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}, substitutep_{i}forp, 4.88-kg for massmof the object and 31.4 m/s for velocity of the object in the equationp=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 momentump_{i}of the object, therefore,

p_{f}= 153 kg. m/s …… (3)In the above figure the angle

θwill be,

θ= 42° + 42°= 84°

Using cosine law, the magnitude of change of the linear momentum of the object Δ

pwill be,Δ

p= √(p_{i})^{2}+ (p_{f})^{2}+ 2 (-p_{i})(p_{f})cosθ…… (4)To obtain magnitude of linear momentum Δ

p, substitute 153 kg. m/s forp_{i}, 153 kg. m/s forp_{f}and 84° for the angle θ in the equation Δp= √(p_{i})^{2}+ (p_{f})^{2}+ 2 (-p_{i})(p_{f}) cosθ,Δ

p= √(p_{i})^{2}+ (p_{f})^{2}+ 2 (-p_{i})(p_{f}) cosθ= √(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.

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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=mvHere,

mis the mass of the body andvis the velocity of the body.The impulse of the net force

Jacting on a particle during a given time interval is equal to the change in momentum Δpof the particle during that interval.So,

J= Δp=

p_{f}-p_{i}Here

p_{f}is the final momentum andp_{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

vand then rebounds with same velocity, thus the change in momentum Δpof the tennis ball will be equal to,Δ

p= (mv) – (-mv)= 2

mvTo obtain the change in momentum Δ

pof the ball, substitute 58-g formand 32 m/s forvin the equation Δp= 2mv, we get,Δ

p= 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= Δ

p/4 msTo obtain the maximum contact force

F_{max}during the collision, substitute 3.7 kg.m/s for Δpin the equationF_{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

Jacting on a particle during a given time interval is equal to the change in momentum Δpof the particle during that interval.

J= Δ_{x}p=

p_{fx}–p_{ix}Here

p_{fx}is the final momentum andp_{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 velocityv_{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= 1/3_{x}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= 1/3_{x}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 forJand 0 m/s for_{x}p_{ix}in the equationp_{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

pis equal to the mass of the bodymtimes velocity of the bodyv.So,

p=mvIn 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 isp_{i,2}, final momentum of the Earth isp_{f,1}and final momentum of the meteorite isp_{f,2}.To obtain the initial momentum

p_{i,1}of the Earth, substitutep_{i,1}forp,m_{1}for the mass of the Earth andv_{1,i}for the initial velocity of the Earth in the equationp=mv,

p_{i,1}=m_{1}v_{1,i}To obtain the initial momentum

p_{i,2}of the meteorite, substitutep_{i,2}forp,m_{2}for the mass of the meteorite andv_{2,i}for the initial velocity of the meteorite in the equationp=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 }= 0So,

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 form_{2}, 5.98×10^{24}kg form_{1}and 7200 m/s forv_{2,I}in the equationv_{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/sFrom 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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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 massm_{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 massm_{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 massm_{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 massm_{1}andv_{2i}is the initial velocity of the body having massm_{2}.When,

m_{1 >> }m_{2}, then,

v_{1f}=v_{1i}andv_{2f}≈ 2v_{1i}-v_{2i}

Solution:-Since the mass of Jupiter

M(m_{1}=M) is very much greater than the mass of the spacecraftm(m_{2}=m), therefore the speed of the spacecraftv_{2f}relative to the Sun will be,

v_{2f}≈ 2v_{1i}-v_{2i}≈ 2

V–v(Since,V=v_{1i}andv=v_{2i})Here, speed of the Jupiter relative to the Sun is

V(=v_{1i}) and speed of the spacecraft relative to the Sun isv(=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 SunVand -12 km/s for speed of the spacecraft relative to the Sunvin the equationv_{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

Jacting on a particle during a given time interval is equal to the change in momentum Δpof the particle during that interval.So,

J= Δp=

p_{f}-p_{i}Here

p_{f}is the final momentum andp_{i}is the initial momentum of the body.Impulse of a force

Jis defined as,

J=F_{av}tHere

F_{av}is the average force andtis the impact time.

J=F_{av}tSo,

F_{av}=J/t

Solution:-The change in momentum Δ

pof one bullet will be,Δ

p=p_{f}-p_{i}=

mv- (-mv)= 2

mvTo obtain the change in momentum Δ

pof one bullet, substitute 3.0 g for massmand 500 m/s forvin the equation Δp= 2mv,Δ

p= 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 impulseJin one minute divided by one minute.So,

F_{av}= 100(J)/t= 100(Δ

p)/tTo find out the average force

F_{av}exerted by the stream of bullets on Superman’s chest, substitute 3.0 kg.m/s for Δpand 60 s fortin the equationF_{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) ifm≥M,showthat 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 massm_{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 massm_{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 massm_{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 massm_{1}andv_{2i}is the initial velocity of the body having massm_{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}andv_{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 area,b, andcfrom left to right.After the first collision between

aandbone has, the velocity ofbwill be,

v_{b}_{,1}=v_{0}and

the velocity of

awill be,

v_{a}_{,1}= 0After the first collision between

bandc, the velocity ofcwill be,

v_{c}_{,1}=2mv_{0}/m+Mand

the velocity of

bwill be,

v_{b}_{,2}=(m-M)v_{0}/(m+M)(a) If

m≥Mthen the ballbcontinue to move to right (or stops) and there are no more collisions.(b) If

m<Mthen ballbbounces back and strikes ballawhich 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

adue to friction is defined as,

a=µ_{k}gHere

µ_{k}is the coefficient of kinetic friction andgis the free fall acceleration.Time

tis equal to the speedvof the object divided by the accelerationaof the object.

t=v/aIn accordance to equation of motion, the distance

dtraveled by the object is equal to,

d=ut+1/2at^{2}Here

uis the initial velocity,tis the time andais the acceleration.As the initial velocity

u=0, thus the equationd=ut+1/2atwill become,^{2}

d=ut+1/2at^{2}= (0)

t+1/2at^{2}= 1/2

at^{2}

Solution:-

(a)For an object with initial speedvand deceleration –awhich travels a distancexbefore stopping.So the time

tto stop will be,

t=v/aThe average speed while stopping is

v/2.The distance

xwill be,

x=1/2at^{2}To obtain the speed

v, substitutev/afortin the equationd=1/2at^{2}, we get,

x=1/2at^{2}= 1/2

a(v/a)^{2}2

x=v^{2}/aOr,

v=√2axTo find out speed in terms of coefficient friction, Substitute

µ_{k}gforain the equationv=√2ax,

v=√2ax=√2 (

µ_{k}g)x=√2

µ_{k}g xTo obtain the speed of car

Aafter the collision, substitute 0.130 forµ_{k}, 9.81m/s^{2}forgand 8.20 m forxin the equationv=√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

Bafter the collision, substitute 0.130 forµ_{k}, 9.81 m/s^{2}forgand 6.10 m forxin the equationv=√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

Aafter the collision will be 4.57 m/s while the speed of carBwill be 3.94 m/s.

(b)Ifv_{0}is the at which carBstruck carA, 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

mis the mass of car_{A}A,mis the mass of car_{B}B,vis the speed of car_{A}Aandvis the speed of car_{B}B.To obtain the speed

v_{0}at which carBstruck carA, substitute 1100 kg form, 4.57 m/s for_{A}v, 1400 kg for_{A}mand 3.94 m/s for_{B}vin the equation_{B}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 carBstruck carAwould 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 distanceytravelled by the body divided by total timettaken by the body to travel that distance.So,

v_{av}=y/tThus the time

twill be,

t=y/v_{av}Momentum of a body (

p) is defined as,

p=mvHere,

mis the mass of the body andvis the velocity of the body.Jacting on a particle during a given time interval is equal to the change in momentum Δpof the particle during that interval.So,

J= Δp=

mΔv(Since, Δp=mΔv)Impulse of a force

Jis defined as,

J=F_{av}tHere

F_{av}is the average force andtis the impact time.

J=F_{av}tSo,

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/sTo obtain the time duration

tof the chop, substitute 2.8 cm foryand 4.8 m/s forv_{av}in the equationt=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

tof the chop will be 5.8 ms.(b) To find out the applied average forceF_{av}, first we have to find out the impulse of forceJ.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 Δ

vin the equationJ=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 forJand 5.8 ms fortin the equationF_{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 andv_{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 andv_{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 manwill 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}= 0So,

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 stonev_{s}and 195-lb for mass of the manm_{m}in the equationv_{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.

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