Problem: The figure shows a bullet of mass 200 g traveling horizontally towards

Question

Problem: The figure shows a bullet of mass 200 g traveling horizontally towards the east with speed 400 m/s, which strikes a block mass 1.5 kg that is initially at rest on a frictionless table. After striking the block, the bullet is embedded in the block, and the block and the bullet move together as one unit. Take east to be the postitive direction

Q1. What is the velocity , in meters per second, of the block/bullet combination immediately after the impact?

Q2. What is the impulse, in newtons-seconds, by the block on the bullet?

Q3. What is the impulse, in newtons-seconds,from the bullet on the block?

Q4. If it took 3 ms for the bullet to change its speed from 400 m/s to its final speed after the impact, what is the average force, in newtons, between the block and the bullet during this time?

Explanation / Answer

Given

mass of bullet m = 0.2 kg

mass of block is M = 1.5 kg

speed of bullet before collision is u1 = 400 m/s

after the collision the bullet embeded in the block and both are moving with speed v

rom conservation of momentum  

total momentum before collision = momentum after the collision

m1*u1+ m2*u2 = (m1+m2)V

0.2*400 + 1.5 *0 = (0.2+1.5) V

V= 45 .07 m/s

Q1 , v = 45.07 m/s

Q2 impulse = change in momentum

m1(v-u1) + m2(v-u2)

0.2(45.07 -400) + 1.5(45.07 -0) = -3.381 N.s

Q3 What is the impulse, in newtons-seconds,from the bullet on the block is

m1(V-u1) = 0.2(400-45.07) = 70.986 N.s

Q4

` we know that the impulse I = F*t

70.986 = F*3*10^-3

F = 23662 N

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