Question
A ball moving with velocity $$2\,m{s^{ - 1}}$$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in $$m{s^{ - 1}}$$ ) after collision will be
A.
0, 1
B.
1, 1
C.
1, 0.5
D.
0, 2
Answer :
0, 1
Solution :
If two bodies collide head on with coefficient of restitution
$$e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\,......\left( {\text{i}} \right)$$
From, the law of conservation of linear momentum
$$\eqalign{
& {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} \cr
& \Rightarrow {v_1} = \left[ {\frac{{{m_1} - e{m_2}}}{{{m_1} + {m_2}}}} \right]{u_1} + \left[ {\frac{{\left( {1 + e} \right){m_2}}}{{{m_1} + {m_2}}}} \right]{u_2} \cr
& {\text{Substituting }}{u_1} = 2\,m{s^{ - 1}},{u_2} = 0,{m_1} = m\,{\text{and}}\,{m_2} = 2m,\,e = 0.5 \cr} $$
we get, $${v_1} = \left[ {\frac{{m - m}}{{m + 2m}}} \right] \times 2$$
$$ \Rightarrow {v_1} = 0$$
Similarly, $${v_2} = \left[ {\frac{{\left( {1 + e} \right){m_1}}}{{{m_1} + {m_2}}}} \right]{u_1} + \left[ {\frac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right]{u_2}$$
$$\eqalign{
& = \left[ {\frac{{1.5 \times m}}{{3m}}} \right] \times 2 \cr
& = 1\,m{s^{ - 1}} \cr} $$