Question

A ball of mass m makes a perfectly elastic head-on collision with a second ball, of mass, M, initially at rest. The second ball moves off with half the original speed of the first ball.
a) Express M in terms of m.
b) Determine the fraction of the original kinetic energy retained by the ball of mass m after the collision.

Answer 1

Answer:-A)The Value of M in terms of m is M=5m

Answer:-The fraction of original retained by the ball is 400/9 percent which is 44.44%.

Explanation:

A ball of mass m makes a perfectly elastic head-on collision with a second ball, of mass, M, initially at rest. The second ball moves off with half the original speed of the first ball.

a) Express M in terms of m.

b) Determine the fraction of the original kinetic energy retained by the ball of mass m after the collision.:-

Here we use momentum and energy conservation:-

Mass m ball moves with u and then it collides with M initally at rest the second ball moves with u/2 after collision.

Momentum:- mu=mv+Mu/2 and

Energy:- 1/2mu^2=1/2mv^2+1/2M(u/2)^2.

Solving both equations to get M and v.

we get:-

M=5m and

v=-3u/2.

Answer:-A)The Value of M in terms of m is M=5m

Fraction retained by the ball m is:-

KEf/KEi=v2/u2=9/4(Answer).

Answer:-The fraction of original retained by the ball is 400/9 percent which is 44.44%.

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Answer 2

Given,

One ball mass =m

Initial speed =u

Second ball mass=M at rest

Final speed v = u/2

To find,

• Express M in terms of m.

• The fraction of the original kinetic energy retained by the ball of mass m after the collision.

Solution,

According to momentum and energy conservation:-

Momentum: mu= mv+ Mv'

Momentum mu= mv + Mu/2

Energy: 1/2mu²=1/2mv²+1/2M(u/2)²

Solving both equations to get M and v.

We get:-

M=5m and v=-3u/2

The fraction of the original kinetic energy retained by the ball of mass m after the collision:

KEf/KEi= v2/u2 =9/4.

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