Question

Particle A makes a perfectly elastic collision with another particle B at rest. They fly apart in opposite direction with equal speeds.. If their masses are mA & mB respectively, then:-


a) 2mA=mB
b) 3mA=mB
c) 4mA=mB
d) (3)1/ 2mA=mB

Answer 1

Answer:

$3m_{A}  = m_{B}$

Explanation:

Let initial velocity of particle A is u and final velocity of particle A and B are v.

Since, the collision is elastic.

According to law of conservation of momentum,

$m_{A} (u )+ m_{B} (0) =m_{A} (v) + m_{B} (-v)$

$m_{A} (u ) = (m_{A}  - m_{B} ) (v)$ ...(1)

$\frac{u}{v} = \frac{m_{A} - m_{B} }{m_{A} }$ ...(2)

Now, applying law of conservation of energy we get,

$\frac{1}{2}  m_{A} u^{2} =\frac{1}{2}  m_{A} v^{2} + \frac{1}{2}  m_{B} v^{2}$

$\frac{1}{2}  m_{A} u^{2} =\frac{1}{2}  (m_{A} + m_{B}) v^{2}$ ...(3)

Dividing equation (3) by equation (1) we get,

$\frac{u}{v} = \frac{m_{A} + m_{B}  }{m_{A} - m_{B}}$ ...(4)

Comparing value of $\frac{u}{v}$ we get,

$\frac{m_{A} - m_{B} }{m_{A} } = \frac{m_{A} + m_{B}  }{m_{A} - m_{B}}$

$\frac{m_{A} }{m_{B} } = \frac{1}{3}$

Hence,

$m_{B} = 3m_{A}$

Answer 2

Answer:mark me brainliest if it helps

Explanation: