First slide

The steps in collision

Question

A ball of mass m is released from the top of an inclined plane of inclination $θ$ as shown in figure. It strikes a rigid surface at a distances $\frac{3 h}{4}$ from top elastically. Impulse imparted to ball by the rigid
surface is

Moderate

A

$m \sqrt{\frac{3}{2} gh}$
B

$m \sqrt{3 gh}$
C

$2 m \sqrt{3 gh}$
D

$m \sqrt{6 gh}$

Solution

Loss in PE = Gain in KE
$m g h_{1} = \frac{1}{2} m v^{2}$
$m g \times \frac{3}{4} h = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{3 g h}{2}}$
Now, impulse imparted, $J = 2 m v = 2 m \sqrt{\frac{3 g h}{2}} = m \sqrt{6 g h}$

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