Question

A ball of mass 250 g moving with a speed of 72 kmph is deflected by a batsman by an angle of 120$^\circ$ without changing its initial speed. The impulse imparted to the ball is:

• A. 253 kg m s$^{-1}$
• B. 53 kg m s$^{-1}$
• C. 52 kg m s$^{-1}$
• D. 5 kg m s$^{-1}$

Hint:

Impulse is the change in momentum due to a collision or redirection of motion. Always convert velocity from km/h to m/s before substitution. For deflection problems, use the vector relation of momenta, not simple difference. The direction of impulse is along the change in momentum vector, not the motion.

Answer 1

The Correct Option is D

• Given data: mass $m = 250$ g $= 0.25$ kg, velocity $v = 72$ km/h $= 20$ m/s, and deflection angle $= 120^\circ$.
• Since the speed remains the same but direction changes, impulse depends on the vector change in momentum.
• The initial and final momenta make an angle of $120^\circ$, so the magnitude of impulse is given by: \[ J = \sqrt{p^2 + p^2 - 2p^2\cos(120^\circ)} = 2p\sin(60^\circ) \] • Substituting $p = mv$: \[ J = 2(0.25)(20)\sin(60^\circ) = 5 \text{ N·s (or kg m s$^{-1}$)}. \] • Hence, the impulse imparted to the ball is 5 kg m s$^{-1$}.