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Collision Example II

A small ball ...

Question

A small ball falling vertically downward with constant velocity 4 m/s strikes elastically a massive wedge moving with velocity 4 m/s horizontally as shown. The velocity of the rebound of the ball is

4 \sqrt{2} m / s

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4 \sqrt{3} m / s

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4 m / s

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4 \sqrt{5} m / s

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Solution

The correct option is D $4 \sqrt{5} m / s$
Along normal to inclined plane velocity of wedge is constant,
$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$ and $v_{2} = (\frac{m_{2} - m_{1}}{m_{1} + m_{2})}) u_{2} + \frac{2 m_{1} u_{1}}{m_{1} + m_{2}}$
Substituting $m_{2} = 0$ , we get
$v_{1} = u_{1}$ and $v_{2} = 2 u_{1} - u_{2} = 6 \sqrt{2}$
Along the plane velocity is constant.= $2 \sqrt{2}$
Net velocity is $4 \sqrt{5}$

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A small ball falling vertically downward with constant velocity 4 m/s strikes elastically a massive wedge moving with velocity 4 m/s horizontally as shown. The velocity of the rebound of the ball is

The correct option is D 4√5 m/sAlong normal to inclined plane velocity of wedge is constant, v1=(m1−m2m1+m2)u1+2m2u2m1+m2 and v2=(m2−m1m1+m2))u2+2m1u1m1+m2 Substituting m2=0 , we get v1=u1 and v2=2u1−u2=6√2 Along the plane velocity is constant.= 2√2 Net velocity is 4√5