A bullet of mass 30 ~g moving with 700 ~ms ⁻¹ collides with a block of mass 4 ~kg hanging by a string of length 0.4 ~m . After collision, the block rises to a height of 0.2 ~m . Then, find the velocity of the bullet when it comes out of the block.

Step-by-step solution

Given, mass of bullet, m_b=30 ~g =0.03 ~kg Velocity of bullet, v_b=700 ~ms ⁻¹ Mass of block, m_B=4 ~kg Height upto which block rises, h = 0. 2m Before collision, momentum of bullet =m_b v_b aligned & =0.03 700 & =21 ~kg - ms ⁻¹ aligned Let v₁, v₂ be the velocities of bullet and block after the collision. Using conservation of momentum, 21=0.03 v₁+4 v₂ ...(i) Using conservation of energy for the block, Change in KE = Work done aligned & 1 2 m_B v₂^2=m_B g h & v₂^2=2 g h & v₂= 2 9.8 0.2 & =1.979 ~ms ⁻¹ aligned Puttin