A 1.2 kg block of wood hangs motionless from strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes embedded inside the block. Immediately after the bullet becomes embedded in the block, the block is observed to have a speed of 8.0 m/s. What was the speed of the bullet before it hit the block

speed of the bullet before it hit the block is 200 m/s

Explanation:

given data

mass of block m1 = 1.2 kg

mass of bullet m2 = 50 gram = 0.05 kg

combine speed V = 8.0 m/s

to find out

speed of the bullet before it hit the block

solution

we will apply here conservation of momentum that is

m1 * v1 + m2 * v2 = M * V ... 1

here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i. e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet

put all value in equation 1

m1 * v1 + m2 * v2 = M * V

1.2 * 0 + 0.05 * v2 = (1.2 + 0.05) * 8

solve it we get

v2 = 200 m/s

so speed of the bullet before it hit the block is 200 m/s