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)

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°

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.

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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 = mv

Here, m is the mass of the body and v is the velocity of the body.

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

So, J = F_max (2 ms+6 ms)/2

        = F_max (4 ms)

So, F_max = J/4 ms

              = 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/s²

      = (930 kg. m/s²) (1 N/1 kg. m/s²)

      = 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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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

To obtain the area, we have to use Simpson’s rule.

Using Simpson’s rule, the area will be,

J_x = 1/3 h (f₀ +4f₁+2f₂+4f₃+……+4f₁3 + f₁4)

Here h (=0.2 ms) is the width of each strip.

So, J_x = 1/3 h (f₀ +4f₁+2f₂+4f₃+……+4f₁3 + f₁4)

         = 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/s²/ 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).

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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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 5×10¹⁰ 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.

To obtain the initial momentum p_i,1 of the Earth, substitute p_i,1 for p, m₁ 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 = m₁ v_1,i

To obtain the initial momentum p_i,2 of the meteorite, substitute p_i,2 for p, m₂ 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 = m₂ v_2,i

The total initial momentum of the meteorite and the Earth is

So, the final momentum of the system will be,

p_f = (m₁ + m₂) 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

m₁ v_1,i + m₂ v_2,i = (m₁ + m₂) v_f

v_f = (m₁ v_1,i + m₂ v_2,i) / (m₁ + m₂)

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+ m₂ v_2,i) / (m₁ + m₂)

          = (m₂ v_2,i) / (m₁ + m₂)

v_f = (m₂ v_2,i) / (m₁ + m₂)

   = (5×10¹⁰ kg) (7200 m/s) /(5×10¹⁰ kg)+ (5.98×10²⁴ 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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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 m₁ and the particle having mass m₂ 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.

v_1f = (m₁ – m₂/ m₁ + m₂) v_1i + (2m₂/ m₁ + m₂) v_2i

and the final velocity v_2f of the body having mass m₂ is,

v_2f = (2m₁/ m₁ + m₂) v_1i + (m₂ – m₁/ m₁ + m₂) v_2i

Here, v_1i is the initial velocity of the body having mass m₁ and v_2i is the initial velocity of the body having mass m₂.

Since the mass of Jupiter M (m₁ = M) is very much greater than the mass of the spacecraft m (m₂ = 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).

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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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.  

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

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

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/s²) (1 N/1 kg.m/s²)

     = 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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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 v₀. 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 m₁ and the particle having mass m₂ 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.

v_1f = (m₁ – m₂/ m₁ + m₂) v_1i + (2m₂/ m₁ + m₂) v_2i

and the final velocity v_2f of the body having mass m₂ is,

v_2f = (2m₁/ m₁ + m₂) v_1i + (m₂ – m₁/ m₁ + m₂) v_2i

Here, v_1i is the initial velocity of the body having mass m₁ and v_2i is the initial velocity of the body having mass m₂.

 As the target particle second cart (m₂) is at rest, thus, v_2i = 0.

Thus, v_1f = (m₁ – m₂/ m₁ + m₂) v_1i and v_2f = (2m₁/ m₁ + m₂) v_1i

After the first collision between a and b one has, the velocity of b will be,

v_b,1= v₀

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=2mv₀/m+M

and

the velocity of b will be,

v_b,2=(m-M)v₀/(m+M)

(a) If m ≥ M then the ball b continue to move to right (or stops) and there are no more collisions.