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Calculating Force Based On The Law Of Conservation Of Momentum(1)

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Question 1.

A ball of mass 0.5 kg moving at 10 ms-1 collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

Solution:

From the equation based on the law of conservation of momentum, we have

M1u1 + m2u2 = m1v + m2v

Given m1 = 0.5 kg, u1 = 10 ms-1, m2 = 0.5 kg, u2 = 0, V =?

Therefore,

0.5 x 10 + 0.5 x 0 = 0.5v + 0.5v

5 + 0 = v

The common velocity the two balls move off together with is 5 ms-1

Question 2.

A body of mass 4.2 kg moving with velocity 10 ms-1 due east, hits a stationary body of mass 2.8 kg. If they stick together after collision and move with velocity v due east, calculate the value of v.

Solution:

The law of conservation of momentum equation is

M1u1 + m2u2 = m1v + m2v

4.2 x 10 + 2.8 x 0 = (4.2 + 2.8) v

42 + 0    = 7v

v  =  42/7 = 6

The final and common velocity the two bodies move with after collision is 6.0 ms-1

Question 3.

A body of mass 5 kg moving with a velocity of 10 ms-2 collides with a stationary body of mass 6 kg. If the two bodies stick together and move in the same direction after the collision, calculate their common velocity.

Solution:

The equation from the law of conservation of momentum is

M1u1 + m2u2 = m1v + m2v

5 x 10 + 6 x 0 = 5v + 6v

50 + 0      = 11v

v = 50/11

= 4.55 ms-1

Question 4.

A tractor of mass 5.0 x 103 kg is used to tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms-1 just before the towing rope becomes taut. Calculate

(i)        speed of the tractor immediately the rope becomes taut;

(ii)        loss in kinetic energy of the system just after the car has started moving.

Solution:

(i). By the law of conservation of momentum, immediately the rope becomes taut, the momentum of the system is conserved and expressed as

m1u1 + m2u2 = m1v + m2v

Where m1 = mass of the tractor, u1 = initial velocity of the tractor, m2 = mass of the car, u2 = initial velocity of the car, v = common and final velocity of both the tractor and the car.

Note: since the car was initially at rest, u2 = 0, therefore, the initial momentum of the car m2u2 = 0

Therefore, from the conservation of momentum equation,

m1u1 + 0 = v (m1 + m2)

5 x 103 x 3 = v (5 + 2.5) 103

15 = v x 7.5

v = 15/7.5

= 2

The final common velocity is also the speed of the tractor when the rope becomes taut = 2 ms-1

(ii). Loss in kinetic energy of the system equals

initial kinetic energy of the system - final kinetic energy

i.e. K.E1 - K.E2

K.E1 = initial kinetic energy = initial kinetic energy of the tractor + initial kinetic energy of the car

= K.E1 (tractor) + K.E1 (car)

Since K.E1 (car) = 0, the initial kinetic energy of the system is K.E1 (tractor).

Generally, the kinetic energy of a system is given by the formula

1/2m1u12

Therefore, the initial kinetic energy of the system is

½ x 5 x 103 x 32

= 22.5 x 103 J

The final kinetic energy of the system:

K.E2 = K.E2 (tractor) + K.E2 (car)

=  1/2m1v2 + 1/2m2v2   (both tractor and car have common velocity)

=  1/2v2 (m1 + m2)

= 1/2 x 22 (5 x 103 + 2.5 x 103)

= 2 (5 + 2.5) 103

= 2 (7.5) 103

= 15 x 103 J

Therefore, loss in kinetic energy of the system  = K.E1 - K.E2

= 22.5 x 103 - 15 x 103

= 103 (22.5 - 15)

= 7.5 x 103

= 7.5 x 103 J