Question

1. A bullet of mass m -25.0 g is fired into a stationary block of mass m, -4.00 kg, which is suspended on a rope, as shown below. The bullet is initially traveling with velocity v. - 400 m/s, passes through the block and emerges with a final velocity horizontally Immediately after the impact the block travels upward with a velocity of 2.00 m's and reaches a vertical height, h before coming to rest. Determine the maximum height the block rises to before coming to rest.[3 1. determine the final speed of the bullet (2) c. Is the collision elastic or inelastic, show work. [2]

Answer 1

a) As the blocks moves with velocity 2m/s vertical up so maximum height raise

02 = 22 - 2gh

$h = \frac{2}{9.8}$

2) let final speed of bullet is vf

$\frac{m{v_i}^2}{2}=\frac{m{v_f}^2}{2}+(m+M)gh$

$\frac{0.025\times{400}^2}{2}=\frac{0.025{v_f}^2}{2}+(0.025+4)\times9.8\times 0.204$

$2000-8.05=\frac{0.025{v_f}^2}{2}$

C) Collision is not elastic because some energy is lost against in potential energy.

Inelastic collision.

Thanks for asking.