The man fires an 84 g arrow so that it is moving at 80 m/s when it hits and embeds in a 8.0 kg block resting on ice. Part A What is the velocity of the block and arrow just after the collision?

0.83m/s

Explanation:

According to law of conservation of momentum, the sum of momentum of bodies before collision is equal to the sum of their momentum after collision. After collision, the body moves together with a common velocity (v).

If m1 and m2 are the masses of the objects

u1 and u2 are their initial velocities

v is the final velocity of the objects after collision.

According to law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

Given mass of the arrow(m1) = 84g = 0.084kg

m2 = 8kg

u1 is the initial velocity of the arrow = 80m/s

u2 = initial velocity of the block = 0m/s(since the block is at rest)

v = the velocity of the block and arrow just after the collision

Substituting this values in the formula we will have;

0.084(80)+8(0) = (0.084+8)v

6.72+0= 8.084v

v = 6.72/8.084

v = 0.83m/s

Therefore, the velocity of the block and arrow just after the collision is 0.83m/s

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow