A ball travelling at 6 m/s collides with another ball twice its mass.The first ball due to the collision moves at 3 m/s in the reverses direction. What is the velocity of the ball that was struck?  



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william1941's profile pic

Posted on (Answer #1)

Here we use law of Conservation of linear momentum.

The collision between the two balls is assumed to be perfectly elastic.

Let us take the mass of the lighter ball to be M. It is moving at 6 m/s before it strikes the heavier ball.

After collision the lighter ball moves in the opposite direction at 3 m/s. Let us take the velocity of the second ball after the collision to be V.

Therefore M*6 + 2M* 0 = M*(-3) + 2M*V

=> 6M = -3M + 2MV

=> 9M = 2MV

=> V = 9/2

=> V = 4.5 m/s

But if we let M = 1kg, we see that the total energy in the system initially = 6^2 = 36 J and the total energy finally is 2*4.5^2 + 3^2 = 49.5 J which has actually increased.

So it isn't possible for the smaller ball to strike the larger and bounce back at 3 m/s.

neela's profile pic

Posted on (Answer #2)

Let M be the mass of the first ball , which collides with the second ball of mass 2M.

The velocity of the first ball is 6m/s.

The second ball's velocity is zero as the ball is asumed to be at rest.

We presume the first ball  moves with the a certain  velocity v1 (not 3m/s as given in the problem) after collision.

Assuming the elastic collision, the velocities of the first and second ball after collision should be as below:

v1 ={ (M-2M)/(M+2M}6m/s = - 2m/s. (negative indicates the velocity is in the reverse direction.

v2 = 2M*6/(m+2M) = 4m/s.

Therefore , the second ball moves with a velocity 4 m/s after collision.

There is an internal inconsistency in the given problem. How can a ball reverse its  velocity at 3m/s  which is higher than the velocity after a perfect elastic collision . Normally the energy is dissipated  even in perfect collision.

Hope this helps an inquisitive mind.

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