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ss7ja [257]
3 years ago
7

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before s

triking the floor, and a speed of 6.5 m/s just after bouncing, find the average force of the floor on the ball. Assume that the ball is in contact with the floor for 0.32 s, and that the mass of the ball is 0.42 kg
Physics
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

F = m*a

so find the change of momentum P

Pdown

P=m*v_1=0.42kg*3m/s

Pup

P=m*v_1=0.42kg*6.5m/s

dP = change in P

dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s

dT = 0.32 s

so

F = \frac{dP}{dt}=\frac{3.99Kg*m/s}{0.32s} =12.468 N

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Two hockey players have a total momentum of +200 kg x m/s before a collision (+ is to the right). After the collision, they move
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Total momentum after the collision: +200 kg m/s to the right

Explanation:

We can answer this question by using the law of conservation of momentum, which states that for an isolated system (=no external forces acting on the system), the total momentum is conserved.

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Learn more about momentum:

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