A 10.0-gram bullet is fired horizontally at two blocks at rest on a frictionless
ID: 1494246 • Letter: A
Question
A 10.0-gram bullet is fired horizontally at two blocks at rest on a frictionless table. In Collision I, 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 400m/s. After passing through Block 1 and before it enters Block 2, the speed of the bullet is 50m/s. Please answer the following questions. What is the final velocity of Blackl after the bullet lias 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
m = 10 * 10^-3 Kg
M1 = 1.2 Kg
Vi = 400 m/s
(a)
Speed of bullet after passing through block 1, Vf = 150 m/s
Using Momentum Conservation,
Initial Momentum = Final Momentum
m*vi + M*0 = m*Vf + M*v
10 * 10^-3 * 400 = 10 * 10^-3 * 150 + 1.2 * v
v = 2.08 m/s
Velcoity of block 1 , when bullet has passed through it, v = 2.08 m/s
(b)
F*t = m*v
F*0.2*10^-3 = 10*10^-3 * (400 - 150)
F = 12500 N
(c)
Initial K.E = 1/2 * 10*10^-3 * 400^2 = 800 J
Final K.E = 1/2 * 10*10^-3 * 150^2 + 1/2 * 1.2*2.08^2 = 115.1 J
As K.E is lost in the collision, it is inelastic.
(d)
Magnitude = 10 * 10^-3 * 400 = 4 Kgm/s
Direction = to the right.
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