Question

A block A of mass m moving with a constant velocity v along a smooth horizontal floor collides with another block B of mass 7m and
rebounds with a velocity 2V/5, the velocity of block B after collision is

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Answer 1

initial velocity of block A of mass ‘m’ , $u_A$ = v

initial velocity of block B of mass ‘7m’, $u_B$ = 0

final velocity of block A , $v_A$ = -2v/5

we have to find out final velocity of block B, $v_B$

using law of conservation of linear momentum,

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B$

or, m × v + 7m × 0 = m(-2v/5) + 7m × $v_B$

or, 7mv/5 = 7m$v_B$

or, $v_B$ = v/5

hence, velocity of block B is v/5