Center Of Mass Momentum And Collision Question 3
Question 3 - 30 January - Shift 1
A ball of mass $200 g$ rests on a vertical post of height $20 m$. A bullet of mass $10 g$, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $30 m$ and the bullet at a distance of $120 m$ from the foot of the post. The value of initial velocity of the bullet will be (if $g=10 m / s^{2}$ ):
(1) $120 m / s$
(2) $60 m / s$
(3) $400 m / s$
(4) $360 m / s$
Show Answer
Answer: (4)
Solution:
Formula: Conservation of Momentum
$v_1=\frac{30}{\sqrt{\frac{2 h}{g}}}, v_2=\frac{120}{\sqrt{\frac{2 h}{g}}}$
$(0.01) u=(0.2) \frac{30 \sqrt{g}}{\sqrt{2 h}}+(0.01) \frac{120 \sqrt{g}}{\sqrt{2 h}}$
$u=300+60=360 ms^{-1}$
BETA
