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Physics Question on work, energy and power

A ball of mass mm moving with a constant velocity uu strikes against a ball of same mass at rest. If ee = coefficient of restitution, then what will be the ratio of velocity of two balls after collision ?

A

1−e1+e\frac{1 -e}{1+e}

B

e−1e+1\frac{e -1}{e + 1}

C

1+e1−e\frac{1+e}{1 -e}

D

e+1e−1\frac{e + 1}{e -1}

Answer

1−e1+e\frac{1 -e}{1+e}

Explanation

Solution

Here, m1=m,m2=m,u1=u,u2=0m_1 = m, m_2 = m, u_1 = u, u_2 = 0
Coefficient of restitution, e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}
e=v2−v1u−0e = \frac{v_2 -v_1}{u -0}
or v2−v1=euv_2 - v_1 = eu ....(i)
According to law of conservation of linear momentum,
m1 u1+m2 u2=m1 v1+m2 v2m_1 \, u_1 + m_2 \, u_2 = m_1 \, v_1 + m_2 \, v_2
⇒  mu=m(v1+v2)\Rightarrow \:\: mu = m (v_1 + v_2)
or v1+v2=uv_1 + v_2 = u ...(ii)
Solving equations (i) and (ii) for v1v_1 and v2v_2, we get
v1=u(1−e)2,v2=u(1+e)2v_1 = \frac{u(1 -e)}{2} , v_2 = \frac{u(1 + e)}{2}
∴  v1v2=1−e1+e\therefore \:\: \frac{v_1}{v_2} = \frac{1 -e}{1 +e}