If a force of 40N is applied for 0.2 sec to change the momentum of a volleyball, what is the impulse?
1 answer:
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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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Answer:
∑Fy = 0, because there is no movement, N = m*g*cos (omega)
Explanation:
We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.
If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.
