Two blocks `m_1 and m_2` of masses 2kg and 5kg, respectively, are moving on a smooth plane along a straight line in the same direction, with velocities `10 m*s^(-1) and 3 m*s^(-1)` respectively. The block `m_2` is situated ahead of the block `m_1`. An ideal spring `(k=1120 N*m^(-1)`) is attached to the back of the block `m_2`. Find the compression of the spring when `m_1` collides with `m_2`.

Explanation

Let initial velocities of blocks `m_1 and m_2 be v_1 and v_2` respectively . After collision , the two blocks combine and move with a velocity V [Fig.1.49].

From the law of conservation of momentum,
`m_1v_1+m_2v_2=(m_1+m_2)V`
or,` 2xx10+5xx3=(2+5)V`
or, `V=5 m *s^(-1)`
Suppose the spring is compressed by x . From the law of conservation of energy , we get
total energy before collision = total energy after collision
or, total kinetic energy of the blocks =kinetic energy of the combined blocks +potential energy of the compressed spring
i.e., `1/2 m_1v_1^2+1/2m_2v_2^2=1/2(m_1+m_2)V^2+1/2kx^2`
`or, 2xx(10)^2+5xx(3)^2=7xx(5)^2+1120xx x^2`
`therefore x^2=1/(16)`
or, `x=1/4=0.25 m`.