A golf ball of mass 50 text{ g} placed on a t | vedclass

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Question

(D) The impulse imparted to the ball is equal to the change in its linear momentum.Impulse $= \int F \, dt = 	ext{Area under the } F-t 	ext{ graph}$

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) The impulse imparted to the ball is equal to the change in its linear momentum.Impulse $= \int F \, dt = 	ext{Area under the } F-t 	ext{ graph}$.Given that the force decreases linearly from $F$ to $0$ in time $t = 0.02 	ext{ s}$, the area is a triangle with base $0.02 	ext{ s}$ and height $F$.$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 0.02 	imes F = 0.01F$.The change in momentum is $\Delta p = m(v_f - v_i) = \frac{50}{1000} 	ext{ kg} 	imes (100 	ext{ m/s} - 0) = 0.05 	imes 100 = 5 	ext{ kg m/s}$.Equating impulse and change in momentum: $0.01F = 5$.$F = \frac{5}{0.01} = 500 	ext{ N}$.

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