Question

A ball of mass m is moving with velocity v it strikes the floor and rebouns with the same velocity if n such balls strike the floor per second then force exerted by the balls on the floor is​

Answer 1

Explanation:

A ball of mass (m) is moving with velocity (v). It strikes the floor and rebounds with the same velocity. If a number (n) of such balls strike the floor per second, then what is the force exerted by the balls on the floor?

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Considering single ball and up direction as positive

(initial momentum of the ball) P_i = -mv

(final momentum of the ball after striking)P_f = mv

(change in momentum) dP = P_f - P_i = 2mv.

dP momentum of ball changes in the time interval of 1 second.

(The average change in momentum for time interval of 1 second) <dp/dt >

=dP/time interval = 2mv/1 = 2mv

<dp/dt> = average force exerted by the floor on the ball during time interval of 1 second = 2mv

average force on the floor = -<dp/dt> = -2mv { By Newton’s Third law}

multiply by n for n balls =- 2nmv

We can only find average force from the given data. As balls will exert force only during the time they are in contact with the floor which is very less and in the leftover time there will be zero force in a time interval. Hence, the answer -2nmv is the average force.

Answer 2

Answer :

◈ As per newton's second law of motion, Force is defined as the rate of change of linear momentum$.$

• F = dP/dt

፨ Change in linear momentum :

➝ dp = mΔv

[Final velocity = v (in opposite direction), Initial velocity = -v]

➝ dp = m[v-(-v)]

➝ dp = 2mv

For n such balls, dp = 2nmv

⇒ F = dp/dt

⇒ F = 2nmv/1

⇒ F = 2nmv