Physics impulse and momentum questions?

Question

i'm stuck on the last 3 questions. please help on any u know thanks.
1. A 20g ball of clay is thrown horizontally at 30m/s toward a 1.0kg block sitting at rest on a frictionless surface. The clay hits and sticks to the block.
a. What impulse does the clay exert on the block?
b. What impulse does the block exert on the clay?
c. Does J(block on clay) = -J(clay on block)?

2.A firecracker in a coconut blows the coconut into 3 pieces. Two peices of equal mass fly off south and west, perpendicular to each other , at 20m/s. The 3rd piece has twice the mass as the other two. What are the speed and direction of the 3rd piece.
3.A 10g bullet is fired into a 10kg wod block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0cm across the table. What was the speed of the bullet?

Answer 1

1. Find the momentum of the thrown ball. Since momentum is conserved, the momentum of the ball and block combined is the same as the momentum that the ball originally had. You can use this to find the velocity of the ball and block after sticking. Then you can use this velocity to find the momentum of the ball alone and the block alone after they are stuck. Impulse is equal to the change of momentum.

2. Give each of the smaller pieces a mass m. then the mass of the larger piece is 2m. Find the total momentum for the two smaller pieces. By conservation of momentum the total momentum of the system must be zero after it explodes (it was zero to start with). So the momentum of the big piece plus the momentum of the two small pieces must add up to zero.

3. You need to the coefficient of friction between two pieces of wood. Work your way backwards. What velocity must the bullet and block have to slide that far? Once you have the velocity you can figure out the momentum of the bullet/block combo. Before the bullet struck, all the momentum was in the bullet. how fast must the bullet be going to have this momentum?

Answer 2

you want contraptions on the a lot. i will assume that's µg. yet rather the contraptions cancel out, so as long as they're consistant, that's ok. exchange in momentum is 1µg x 5e7 + 1µg x 0.9 x 5e7 ?P = 1µg x 5e7(a million+0.9) = 50(a million.9) gm/s imparted to the gold, that's ?P = 50(a million.9) = 197µg x V V = 482000 m/s .