Question

Two blocks of masses m1 = 3 kg and m2 = 2 kg rest on a frictionless surface. A bullet of mass m = 0.02 kg strikes block 1 and embeds itself in block 2. Afterwards, block 1 moves with speed v1 = 2 m/s and block 2 (with the bullet inside) moves with speed v2 = 5 m/s. What was the initial speed v0 of the bullet?

Answer 1

Given that block of mass m1 = 3kg and mass m2 = 2kg are present with initial velcities, u1 and u2 equals to 0 as there are at rest.

A bullet of mass m=0.02 kg strikes with the block and embeds into the second block.

After this, speed of block 1 is v1= 2m/s and speed of bullet+block2 is v2= 5m/s

According to conservation of momentum,

`m1* u1+m2* u2+m* u=m1* v1+(m2+m)v2`

Here, u is the initial velocity of bullet,

Substituting the values, we get

`3*0+2*0+0.02*\text{ u = 3}*2+(2+0.02)*5`

`\begin{gathered} u=(16.1)/(0.02) \\ =805\text{ m/s} \end{gathered}`

Thus, the initial speed of bullet is 805 m/s