A 10.0-gram bullet is fired horizontally at two blocks at rest on a friction les

Question

A 10.0-gram bullet is fired horizontally at two blocks at rest on a friction less table. In Collision I, the bullet passes through Block 1 (m i = 1.20 kg). Subsequently, in Collision II the bullet embeds itself in Block 2. The initial speed of the bullet is 400m/s. After passing through Block 1 and before it enters Block 2, the speed of the bullet is 150m/s. Please answer the following questions. What is the final velocity of Block l after the bullet has passed through it? Assume that the bullet required 0.200ms 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

here,

mass of bullet , m = 0.01 kg

mass of block1 , m1 = 1.2 kg

mass of block 2 , m2 = 2 kg

initial speed of bullet , u = 400 m/s

speed of bulllet after collison with block 1 , v1 = 150 m/s

(a)

let the final velocity of block 1 be v2

using conservation of momentum

m * u = m*v1 + m1 * v2

0.01 * 400 = 0.01 * 150 + 1.2*v2

v2 = 2.08 m/s

(b)

t = 0.0002 s

the average force , F = m1 * v2 /t

F = 1.2 * 2.08 /( 0.0002)

F = 12500 N

the average force acting on the block is 12500 N in the direction of bullet velocity

(c)

for collison (1)

initial kinetic energy , KEi = 0.5 * m * u^2

KE = 800 J

final kinetic energy , KEf = 0.5 * m * v1^2 + 0.5 * m1 * v2^2

KEf = 115.1 J

KEf is not equal to KEi

so, the collison is inelastic

(d)

for collison (2)

final velocity of combination be v3

using conservation of momentum

m* v1 = ( m + m2 ) * v3

0.01 * 150 = ( 0.01 + 2) * v3

v3 = 0.746 m/s

the total momentum , P = m1 * v2 + ( m + m2) * v3

P = 4 kg.m/s to the right

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