A plastic block of mass 0.2kg is initially at rest on a horizontal frictionless surface. ?
A plastic block of mass 0.2kg is initially at rest on a horizontal frictionless surface. A bullet of mass 12 g is fired with an initial speed of 550m/s towards the block. The bullet passes completely through the block and exits the other side with a speed of 20.0m/s. What percent of the initial kinetic energy of the bullet+block system is lost during the collision? Give your answer as a positive percentage.
- Anonymous2 months agoFavorite Answer
mb*Vb = mb*V'+(mbl*Vbl)
Vbl = 12/1000*(550-20)/0.2 = 31.8 m/sec
lost energy in % = 100*(0.006*(550^2-20^2)-0,10*31.8^2)/(0.006*550^2) = 94.3
- Anonymous2 months ago
12 g * 550 m/sec = 12 g * 20 m/sec + 200 g * v
solve for the v of the block.
Calculate the KE of the bullet before and after and KE of the block.
% = (KE of the bullet after + KE of the block) / KE of the bullet before * 100