Question

two curling stones collide on an ice rink. stone 1 has a mass of 20kg and is moving north. Stone 2 was at rest initially. the stones collide dead center, sending stone 1 at 0.35 m/s and giving stone 2 a final velocity of 0.75 m/s to the north. Find the mass of stone 2 and the initial velocity of stone 1.

Answer 1

Given:

$m_1=20\:\rm kg;$

$v_2=0\:\rm m/s;$

$v_1'=0.35\:\rm m/s;$

$v_2'=0.75\:\rm m/s.$

The law of conservation of momentum states

$$m_1{\bf v}_1+m_2{\bf v}_2=m_1{\bf v}_1'+m_2{\bf v}_2'.$$

The law of conservation of energy states

$$\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}=\frac{m_1v_1'^2}{2}+\frac{m_2v_2'^2}{2}.$$

In our case, we get equations

$$20v_1=20\times 0.35+m_2\times 0.75.$$

$$20v_1^2=20\times 0.35^2+m_2\times 0.75^2.$$

Solution:

$v_1=0.4\:\rm m/s;$

$m_2=1.3\:\rm kg.$