Question

A 0.45 kg sphere swings down from a height of 0.48 meters and elastically collides with a 0.74 kg block on a horizontal smooth surface as shown above. How far along the 30 degree incline (not vertical height) does the block slide before coming to rest?

Answer 1

Answer on Question 39102, Physics, Mechanics

Elastically colliding means we can use both energy and momentum conservation law here. Speed of sphere before colliding is $v_{s}=\sqrt{2gh}$ and its energy is $m_{s}gh_{s}=\frac{m_{s}v_{s}^{2}}{2}$. From conservation laws

$\frac{m_{s}v_{s}^{2}}{2}=\frac{m_{1}v_{1}^{2}}{2}+\frac{m_{2}v_{2}^{2}}{2}$

$m_{s}v_{s}=m_{1}v_{1}+m_{2}v_{2}$

one can easily find velocity of block after collision

$v_{2}=\frac{v_{s}}{m_{2}/m_{1}+1}$

From this we can find how high will the block get

$h_{2}=\frac{v_{2}^{2}}{2g}$

Knowing the height, we will find how far does the block slide

$l=\frac{h_{2}}{\sin 30{}^{\circ}}$

Gathering everything together

$l=\frac{\frac{v_{2}^{2}}{2g}}{\sin 30{}^{\circ}}=\frac{\left(\frac{v_{s}}{m_{2}/m_{1}+1}\right)^{2}}{2g\sin 30{}^{\circ}}=\frac{\left(\frac{\sqrt{2gh}}{m_{2}/m_{1}+1}\right)^{2}}{2g\sin 30{}^{\circ}}\approx 0.069\,m=6.9\,cm$