. A bullet of mass 0.01 ~kg travelling at a speed of 500 ~ms ⁻¹ strikes a block of mass 2 ~kg which is suspended by a string of length 5 ~m . The centre of gravity of the block is found to rise a vertical distance of 0.1 ~m . What is the speed of the bullet after it emerges from the block?

Step-by-step solution

According to question, the given situation is shown in the following figure. Mass of block, M=2 ~kg Mass of bullet, m₁=0.01 ~kg Speed of buIlet, v=500 ~ms ⁻¹ h=0.1 ~m If v₁ and v₂ be the velocities of the bullet and blocks after collision, then by conservation of energy, aligned 1 2 M v₂^2 & =M g h v₂ & = 2 g h = 2 9.8 0.1 =1.4 ~ms ⁻¹ aligned According to the law of conservation of momentum, aligned m₁ v & =m₁ v₁+M v₂ v₁ & = m₁ v-M v₂ m₁ = 0.01 500-2 1.4 0.01 & = 5-28 0.01 = 2.2 0.01 =220 ~ms ⁻¹ aligned