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

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