Understanding Force

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

 

Force Lessons

Newton's First Law of Motion
Newton's Second Law of Motion
Newton's Third Law of Motion
Weight and Mass
Force
Impulse
Momentum
Calculating Force
Friction

  

 

 

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