Question

A 100 g body moving with velocity v strikes elastically another body at rest and
continues along the same direction with
velocity v/4. The mass of the other body is
Ops: A
250
B 30g
C 60g
D 75g​

Answer 1

Explanation:

300 g is the answer ( not related to options)

As

m1v1 = m2v2

100×v = (100+x)×v/4

From here

x= 300g

Answer 2

Answer:

The mass of the other body, m₂, calculated is $60g$.

Therefore, option c) $60g$ is correct.

Explanation:

Data given,

The mass of the first body, m₁ = $100g$

The initial velocity of the first body, v₁ = v

After elastically striking, the final velocity of the first body, v₂ = $\frac{v}{4}$

The initial velocity of the second body, u₁ = $0$

After striking, the final velocity of the second body, u₂ = u

The mass of the second body, m₂ =?

Now, we can apply the conservation of linear momentum:

• $m_{1}v_{1} + m_{2}u_{1} = m_{1}v_{2} + m_{2}u_{2}$
• $100v + m_{2} (0) = 100(\frac{v}{4}) + m_{2}u$
• $100v = 25v + m_{2}u$
• $75v = m_{2}u$  
• $v = \frac{ m_{2}u}{75}$-------equation (1)

Also, there is no loss of energy; we can apply the conservation of energy:

• $\frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}u_{1}^{2} = \frac{1}{2} m_{1}v_{2}^{2} + \frac{1}{2} m_{2}u_{2}^{2}$
• $\frac{1}{2} 100v^{2} + \frac{1}{2} m_{2} (0)^{2} = \frac{1}{2} 100(\frac{v}{4})^{2} + \frac{1}{2} m_{2}u^{2}$
• $50v^{2} = 50(\frac{v^{2} }{16}) + \frac{1}{2} m_{2}u^{2}$
• $50v^{2} =\frac{25v^{2} }{8} + \frac{1}{2} m_{2}u^{2}$
• $50v^{2} -\frac{25v^{2} }{8} = \frac{1}{2} m_{2}u^{2}$
• $\frac{400v^{2} - 25v^{2} }{8} = \frac{1}{2} m_{2}u^{2}$
• $\frac{375v^{2} }{8} = \frac{1}{2} m_{2}u^{2}$
• $\frac{375v^{2} }{4} = m_{2}u^{2}$

Now, after putting the calculated value of v in the equation, we get:

• $\frac{375 }{4} ( \frac{m_{2}u }{75})^{2} = m_{2}u^{2}$
• $\frac{375 }{4} \times \frac{m_{2}^{2} u^{2} }{5625} = m_{2}u^{2}$
• $375m_{2} =22500$
• $m_{2} =60g$

Hence, the mass of the other body, m₂, calculated is $60g$.