Question

A smooth ball of mass $1$ kg is projected with velocity $7m/s$ horizontal from a tower of height $3.5m$. It collides elastically with a wedge of mass $3$ kg and inclination of 45${}^o$ kept on ground. The ball collides with the wedge at a height of $1m$ above the ground. Then,

• A. Velocity of the wedge after collision is $5\sqrt{2}$ m/s
• B. Velocity of the wedge after collision is $4$ m/s
• C. Velocity of the the ball after collision is $5\sqrt{2}$ m/s
• D. Velocity of the the ball after collision is $4$ m/s

Answer 1

Answer: (B), (C)

The correct options are
B Velocity of the wedge after collision is $4$ m/s
C Velocity of the the ball after collision is $5\sqrt{2}$ m/s

As the ball is thrown horizontally its going to have only horizontal velocity

$v_x=7m/s$

and the vertical velocity in projectile motion is

$v^2=u^2+2gh$

initial velocity is zero in vertical direction therefore u= 0

$v^2=0+2gh$

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 2.5}=7m/s$

$v_x=v_y$ so it strikes the plane perpendicularly.

Let the ball rebounds with velocity $v$ and and wedge $v_1$

Appy conservation of momentum in linear direction

$m_1{v}_{1,i}+m_2{v}_{2,i}=m_1{v}_{1,f}+m_2{v}_{2,f}$

$1\times 7=-1\times \frac{v}{\sqrt{2}}+3v_1$

$7\sqrt{2}=-v+3\sqrt{2}{v}_1$ ------------( i )

$e=\frac{\frac{v}{\sqrt{2}}+v_1}{7\sqrt{2}}$

$1\times 7\sqrt{2}=\frac{v_1}{\sqrt{2}}+v$..................( ii )

add eq (i) and (ii)

$14\sqrt{2}=v_1(\frac{1}{\sqrt{2}}+3\sqrt{2})$

$v_1=\frac{14\sqrt{2}\times \sqrt{2}}{7}=4m/s$

$7\sqrt{2}=-v+3\sqrt{2}\times 4$

$v=5\sqrt{2}m/s$