A 10.0-grams bullet is fired horizontally at two blocks at rest on a frictionles

Question

A 10.0-grams bullet is fired horizontally at two blocks at rest on a frictionless table. In collision 1, the bullet passes through Block 1 (m_1 = 1.20 kg). Subsequently, in collision II the bullet embeds itself in Block 2. The initial speed of the bullet is 400 m/s. After passing through Block 1 and before it enters Block 2, the speed of the bullet is 150 m/s. Please answer the following questions. What is the final velocity of Block 1 after the bullet has passed through it? Assume that the bullet required 0.200 ms to pass through Block 1. Calculate the magnitude and direction of the average force on the bullet by Block 1. Is collision I between the bullet and Block 1 elastic or inelastic? Provide a brief explanation for your answer. What are the magnitude and direction of the total momentum of the system composed of the bullet and the two blocks?

Explanation / Answer

mass of the bullet,m1=10 gm

mass of block,m2=1.2 kg

mass of second block,m3=2kg

initila speed of bullet,v1=400 m/s

speed of bullet in second block,v2=150 m/s

a) The collision with block 2 is totally inelastic (so Kinetic Energy is not conserved).

mbullet*vbullet = (mbullet+m3)*v2

0.001*150/ (0.001 + 2)=v3,velocity of second block = 0.0749 m/s

mbullet*400 = mbullet*400 + m2*v4

v4 = 2(0.001*400)/1.2 = 0.667 m/s

b)Force is acting by the block on bullet, So, force, F=mv/t=1.2*0.667/0.2*10-3s=4000 N

c)It is an elastic collission since the energy and momentom is conserved during the collision

d)Total momentom is, m1v1+m2v3+m3v4=1.35 kg.m/s

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