Question

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms$^{-1}$. If the mass of the ball is 0.15 kg, determine the change in momentum of the ball. (Assume linear motion of the ball)
(A) 3.6 kg m/s
(B) 5 kg m/s
(C) 3.5 kg m/s
(D) 7 kg m/s

Answer 1

Hint: Momentum is defined as the product of mass and velocity. For the given case, the momentum will change because the direction of velocity has changed after the collision. So basically approaching and receding momentum will add due to direction change.

Complete answer:
Let the momentum of the incoming ball be defined as:
${p_1} = mv$
Where m is the mass of the ball and v is the velocity of the incoming ball.
The momentum of the ball leaving after hitting the bat will be:
${p_2} = - mv$,
Here, we can see that the direction only has simply changed as it was given to us that the speed of the ball has not changed after hitting the bat.
The change in momentum of incoming and outgoing ball is given as:
$\Delta p = {p_1} - {p_2}$.
Substituting the values in this we get:
$\eqalign{
  & \Delta p = mv - ( - mv) = 2mv \cr
  & \Rightarrow \Delta p = 2 \times 0.15 \times 12{\rm{ = 3}}{\rm{.6 kg m}}{{\rm{s}}^{{\rm{ - 1}}}} \cr} $

Therefore the correct option is (A).

Additional Information:
In the small time when the bat hits the ball, we say that an impulse acts as momentum changes in a very short time. Such short timed impacts fall under the category of impulse.

Note:
One might think that as the velocity is not changing, the magnitude of momentum is not changing but that is not the whole story. The momentum is a vector quantity so direction plays an important role. If we do not reverse the sign of velocity, the result will be zero. When handling vector quantities we always set a positive and negative direction according to our convenience. Here we would have chosen the incoming direction as negative, the magnitude of the resultant would have not changed.