Question

a ll #6 m)= 1.8 4l Hot lon bes t bulles

Basically a block is dropped from a building and underneath at the exact same time a bullet is fired and when the bullet hits the block it embeds in the block and brings it up to its original spot and she wants to know how much time it takes for the block to return to its original spot

Answer 1

Let m1,m2 be the masses annd v1,v2 be the initial velocities of theblock and bullet respectively. Suppose the block+bullet move with a velocity V after collision. Taking vertically downward as the positive direction and using law of conservation of momentum, we have,

Initial momentum = Final momentum

=> m1v1 - m2v2 = (m1+m2) V

=> V = m1v1 - m2v2/(m1 + m2)

=> V = 1.8*0 - 0.015*810/ (1.8+0.015)

= - 6.69 m/s

The negative sign shows that the block+bullet system move in vertically upward direction after collision. When the block comes back to the original position its velocity becomes zero at that position. Also the block+bullet move under the effectof gravity after collision, so, acceleration of the system, a = -g = -9.81 m/s^2

If t be the required time, then using, a = v - u/t, where, v = 0, u = V = -6.69 m/s and a = -9.81 m/s^2

Thus, t = v - u/a = (0 - V)/(-g) = V/g = 6.69/9.81

Therefore, time required by the block to come back to the original position after collision with the bullet, t = 0.68s