An 8.08 kg mass moving east at 16.3 m/s on a frictionless horizontal surface col

Question

An 8.08 kg mass moving east at 16.3 m/s on a frictionless horizontal surface collides with a 11.8 kg mass that is initially at rest. After the collision, the 8.08 kg mass moves south at 2.38 m/s. What is the magnitude of the velocity of the 11.8 kg mass after the collision?

What is the angle relative to the east?

What percentage of the initial kinetic energy is lost in the collision?

2.

The bullet remains in the block, and after the impact the block lands d = 1.82 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?

What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.

An 8.35g bullet is fired into a 315g block that is initially at rest at the edge of a table of h = 1.01 m height (see the figure below).

Explanation / Answer

initally let the vel of the bullet be v1 and combined vel of block and bullet be v
using conservation of momentum
mv1=(m+M)v (where m=mass of bullet, and M=mass of block)
8.35v1=(8.35+0.315)v2
8.35v1=8.665v2 -(1)
now height of the block h=1.01m and initail vel along vertical directn is 0m/s
thus h=(gt^2)/2
1.01=4.9t^2
t=0.45 sec
again with the vel of the block due to its collision with the bullet the will be a horizontal vel which will result in the displacement d
d=vt
1.82=v*0.45
v=4.04m/s
now v1=(8.665*0.45)/8.35

v1=4.19m/s
thus vel of bullet = 4.19m/s and vel of block=4.04m/s

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