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# Worked examples on dynamics

Example 1

A ball of mass m. It is dropped onto a horizontal plate as shown in the figure below

Just has the ball makes contact with the plate, it has a velocity v, momentum p and kinetic energy Ek.

1. Write down the momentum p in terms of mass m and velocity v
2. Show that Ek =p2/2m

iii.                  Just before the impact with the plate, the ball of mas 35g has speed 4.5ms-1. It bounces from the plate so that its speed immediately after losing contact with the plate is 3.5ms-1

. The ball is in contact with the plate for 0.14s. Calculate for the time that the ball is in contact with the plate, the average force, in the addition to the weight of the ball, that the plate exert on the ball.

Solution

i   p = mv

ii   Ek = mv2/2    ………..   1

p = mv     ………………    2

eqn1 can be written as

E_k=(m^2 v^2)/2m

therefore

Ek =p2/2m

iii.

Change in momentum ∆p= m(u+v)
The reason for positive sign is because the ball moves in opposite direction after rebound
i.e ∆p = 0.035(3.5+4.5)

∆p=0.28〖kgms〗^(-1)
F= ∆p/∆t
F is the force of impact
F=0.28/0.14

F = 2N

Example 2

A ball falls vertically onto horizontal ground and rebounds, as shown

The ball has momentum p1 downwards just before hitting the ground. After rebounding, the ball

leaves the ground with momentum p2 upwards. The ball is in contact with the ground for 0.020 s.

During this time interval, an average resultant force of 25 N acts on the ball.

What is a possible combination of values for p1 and p2?

Solution

F= (p2- p1)/∆t
Since the ball is in opposite direction, it implies that
F= (p2+p1)/∆t

p2+p1 = 25* 0.02 = 0.5
Since the ball has the same mass the velocity before hitting the plate will be higher than the velocity after it bounces. So therefore p1 > p2, from the option C is the correct answer

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