Question

26. A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest
but free to move. At what speed does the target move off?​

Answer 1

A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move.

From above data we have; mass of bullet is 10 g or 0.01 kg, velocity is 200 m/s and mass of target is 2 kg.

Total mass = (0.01 + 2)kg = 2.01 kg

We have to find the speed with which the target move off.

According to law of conservation of momentum,

Total momentum before collision is equal to total momentum after collision.

m1 × v1 = m2 × v2

Substitute the value of m1 = 0.01, v1 = 200 and m2 = 2.01

→ 0.01 × 200 = 2.01 × v2

→ 2 = 2.01 × v2

→ 2/2.01 = v2

→ 0.99 = v2

Therefore, 0.99 m/s is the speed of the target with which it move off.

Answer 2

Initial mass of bullet , m1 = 10 g = 0.01 kg

Initial velocity of bullet , v1 = 200 m/s

Final mass = mass of bullet + mass of target

⇒ Final mass , m2  = 0.01 + 2 = 2.01 kg

Final velocity of target , v2 = ? m/s

Here we need to apply the conservation of momentum .

$\implies \bold{m_1.v_1=m_2.v_2}\\\\\implies \bold{0.01*200=2.01*v_2}\\\\\implies \bold{v_2=\frac{2}{2.01} }\\\\\implies \bold{\bf{\red{v_2=0.99\;m/s}}}$

So the target will move off with 0.99 m/s