Question

# A mass of $$\displaystyle { m }_{ 1 }$$ is a moving with velocity $$v$$. It collides head on elastically with a stationary ball of mass $$\displaystyle { m }_{ 2 }$$. The velocity of ball becomes $$\displaystyle \frac { v }{ 3 }$$ after collision,then the value of the ratio $$\displaystyle \frac { { m }_{ 2 } }{ { m }_{ 1 } }$$ is:

A
0.5
B
2
C
3
D
4

Solution

## The correct option is A $$0.5$$Given :       $$u_1 = v$$               $$u_2 = 0$$         $$v_1 = \dfrac{v}{3}$$              $$e =1$$Using        $$v_1 = \dfrac{(m_1 - em_2) u_1 + (1 + e) m_2 u_2}{m_1 + m_2}$$$$\dfrac{v}{3} = \dfrac{(m_1 - 1 \times m_2) v + (1 + 1) m_2 (0)}{m_1 + m_2}$$$$\dfrac{1}{3} = \dfrac{(m_1 - m_2) }{m_1 + m_2}$$OR     $$m_1 + m_2 =3m_1 - 3m_2$$        OR       $$2m_1 = 4 m_2$$                   $$\implies \dfrac{m_2 }{m_1 } =\dfrac{1}{2}$$Physics

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