An object with total mass m total = 7.7 kg is sitting at rest when it explodes i
ID: 1786845 • Letter: A
Question
An object with total mass mtotal = 7.7 kg is sitting at rest when it explodes into two pieces. The two pieces, after the explosion, have masses of m and 3m. During the explosion, the pieces are given a total energy of E = 42 J.
1)
What is the speed of the smaller piece after the collision?
m/s
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2)
What is the speed of the larger piece after the collision?
m/s
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3)
If the explosion lasted for a time t = 0.02 s, what was the average force on the larger piece?
N
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4)
What is the magnitude of the change in momentum of the smaller piece?
kg-m/s
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5)
What is the magnitude of the velocity of the center of mass of the pieces after the collision?
Explanation / Answer
The momentum before the explosion = 0 so the momentum after the explosion is also = 0
mv + 3mV = 0
By the energy given
0.5mv^2 + 0.5(3m)V^2 = 42
solving for V in the momentum equation
V = -mv/3m = -v/3
Plugging into the energy equation
0.5mv^2 + 0.5(3m)(-v/3)^2 = 42
0.5mv^2 + (1.5/9)mv^2 = 42
v^2 = 42/(2/3)m
4m = 7.7 kg so m = 1.925 kg and 3m = 5.775
v^2 = 63/1.925
v = 5.72 m/s .............(m)
v = 3.30 m/s ...........(3m)
Ft = (3m)V
V = 3v = 3(5.72) = 17.16 m/s
F(0.02) = 3(1.925)(17.16)
F = 4954.95 N
change in momentum for smaller>
1.925(5.72) = 11.011 kg m/s
velocity of the center of mass = 0
Again by conservation of momentum
momentum of the system is always 0 the momentum for a system of particles is the total mass of the system
times the velocity of the center of mass, since there is mass velocity of center of mass = 0 m/s