# A ball of mass m1 = 8.0 × 10−2 kg starts from rest and falls vertically downward from a height of 3.0 m. After colliding with the ground, it bounces up to a height of 2.0 m. The collision...

A ball of mass m1 = 8.0 × 10−2 kg starts from rest and falls vertically downward from a height of 3.0 m. After colliding with the ground, it bounces up to a height of 2.0 m. The collision takes place over a time interval of t = 5.0 × 10−3 s.

How do I calculate

(i) the momentum of the ball immediately before and immediately after the collision,

(ii) average force exerted by the ground on the ball and,

(iii) impulse imparted to the ball?

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B) and C)

The force exerted by the ground on the ball and the impulse imparted to the ball are determined by the equation

`vecp_f - vecp_i = vecF*Deltat` , where `vecp_i` is the momentum before the collision, `vecp_f` is the momentum before the collision, and `Delta t` is time interval over which the collision takes place.

The initial momentum is directed **downward**, and its magnitude, as found in the above answer, is `p_i = 0.61 kg*m/s` .

The final momentum is directed **upward** and its magnitude, as found in the answer above, is `p_(f) = 0.5 kg*m/s` .

The **vector difference** between `vecp_f ` and ` ` `vecp_i` is then directed **upward **and its magnitude is the **sum** of the magnitudes of the momenta:

`p_f + p_i = 0.61 + 0.5 = 1.11 kg*m/s`

This means that the force exerted from the ground on the ball is also directed **upward**. The magnitude of the impulse imparted on the ball during the collision is

`F*Deltat = 1.11 kg*m/s`

and the magnitude of the force is

`F = 1.1/(5*10^(-3)) = 222 N`

**The impulse is 1.11 kg*m/s and the force is 222 N.**

First we have to calculate the velocity of ball immediately before collision and immediately after it.

We can use the equation of motion: v^2 = u^2 + 2as

where, v is the final velocity, u is the initial velocity, a is acceleration (acceleration due to gravity in this case) and s is the distance traveled.

Before collision, s = 3 m, u = 0 m/s and a = g = 9.81 m/s^2

Thus, v^2 = 2 x 9.81 x 3

solving this, we get, v = 7.67 m/s

Thus, momentum immediately before collision = mv = 8 x 10^-2 x 7.67 kg m/s

= **0.614 kg m/s **

Similarly, after the collision, ball rises to a height of 2 m.

Thus, s = 2 m, v = 0, a = -g = -9.81 m/s^2

Thus, 0 = u^2 + 2 x (-9.81) x 2

solving this, we get, u = 6.26 m/s

Thus, momentum immediately after collision = mu = 8 x 10^-2 x 6.26 kg m/s

= **0.501 kg m/s**

b) Force exerted can be calculated as:

F = m(v-u)/t = 8 x 10^-2 x (7.67 - 6.26)/(5 x 10^-3) = **22.6 N**

c) Impulse = Ft = 22.6 x 5 x 10^-3 = **0.113 kg m/s**

Hope this helps.

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