An object with total mass m total = 7.7 kg is sitting at rest when it explodes i

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

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