Question

A batsman hits back a ball of mass 0.15 kg straight in the direction of the bowler without changing its initial speed of $12m{s^{ - 1}}$. If the ball moves linearly, then the impulse imparted to the ball is
A. 2.8 Ns
B. 3.6 Ns
C. 3.9 Ns
D. 4.2 Ns

Answer 1

Hint: When an object of some mass moves with some velocity then it has kinetic energy and momentum. Now due to some impact if the velocity of that object changes then we can tell that its kinetic energy has changed and momentum has also changed. Now the rate of change of that momentum is a force while change in that momentum is called an impulse.

Formula used:
$I = \Delta p$

Complete answer:
Let us assume a body of mass(m) moving with the velocity(v) then it will be having momentum(p). Momentum is a vector which means it has the direction and magnitude. That momentum is the product of the mass and the velocity. Direction of momentum is nothing but the direction of velocity of the body. Its magnitude will be the product of its mass and its velocity. Here mass of the ball is given as 0.15kg and the initial velocity of the ball before hitting the bat is $12m{s^{ - 1}}$. After hitting the ball it was given that the magnitude of velocity is the same but the direction of velocity is reversed.
Momentum $p = mv$
Initial momentum ${p_i} = 0.15 \times 12 = 1.8Ns$
Final momentum ${p_f} = - 0.15 \times 12 = - 1.8Ns$
Negative sign indicates that the velocity of the ball is reversed.
So impulse is change in momentum which will be $I = \Delta p = 1.8 - ( - 1.8) = 3.6Ns$

Hence answer would be option B.

Note:
Change in momentum is called as impulse but if that change happens very fast i.e within micro second or milliseconds then rate of change of momentum will be very high which means that the force will be very high and that kind of forces will be called as impulsive forces. When collision happens we don’t consider gravitational force as an external impulsive force so we conserve the momentum.