Question

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it elastically strikes another block that is only one-half its mass. Find the velocity of block M at the bottom of the incline before the collision with the small block m.

Answer 1

Given :

M = 0.35 kg

`$m=(M)/(2)=0.175 \ kg$`

Total mechanical energy = constant

or
`$K.E._(top)+P.E._(top) = K.E._(bottom)+P.E._(bottom)$`

But
`$K.E._(top) = 0$` and
`$P.E._(bottom) = 0$`

Therefore, potential energy at the top = kinetic energy at the bottom

`$\Rightarrow mgh = (1)/(2)mv^2$`

`$\Rightarrow v = √(2gh)$`

`$=√(2 * 9.8 * 0.35)$` (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

`$v_1=(m_1-m_2)/(m_1+m_2)v_1+ (2m_2v_2)/(m_1+m_2)$`

here,
`$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$`

∴
`$v_1=(0.35-0.175)/(0.5250)+(2 * 0.175 * 0)/(0.525)`

`$=(0.175)/(0.525)+0$`

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s