A bullet (m1=25 g) fired with a velocity 400 m/s gets embedded into a bag of sand (m2=4.9 kg) suspended by a rope. The velocity gained by the bag is nearly:

Question

A bullet (m1=25 g) fired with a velocity 400 m/s gets embedded into a bag of sand (m2=4.9 kg) suspended by a rope. The velocity gained by the bag is nearly: (A) 2 m/s (B) 4 m/s (C) 8 m/s (D) 0.2 m/s

Tardigrade Admin · Accepted Answer

Linear momentum of a system is conserved in absence of any external force. Conservation of linear momentum gives

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

Given,

m_{1} = 25 g, u_{1} = 400 m / s

m_{2} = 4.9 k g, u_{2} = 0, v = ....

Hence,

\frac{25}{1000} \times 400 + 4.9 \times 0 = (\frac{25}{1000} + 4.9) v

or

v = \frac{10000}{4925} \approx 2 m / s

Note: Since, bullet is embedded into bag, hence the collision is inelastic. But still the momentum is conserved.

Q. A bullet (m1​=25g) fired with a velocity 400 m/s gets embedded into a bag of sand (m2​=4.9kg) suspended by a rope. The velocity gained by the bag is nearly:

2 m/s

4 m/s

8 m/s

0.2 m/s

Q. A bullet $(m_{1} = 25 g)$ fired with a velocity 400 m/s gets embedded into a bag of sand $(m_{2} = 4.9 k g)$ suspended by a rope. The velocity gained by the bag is nearly: