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A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the sp
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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