Question

An object of mass 4 kg is moving in a straight line with kinetic energy 106.58 J. A force is applied in the direction of its motion for 2.7 seconds, and as a result, its kinetic energy is multiplied by a factor of 1.78. a) By what factor is its momentum multiplied? b) What was the magnitude of the force applied to the object? c) Instead of this force, if a force of the same magnitude is applied to the object for the same duration, but is perpendicular to the direction in which the object was originally moving, by what factor is the kinetic energy multiplied?

Answer 1

1)

LONG METHOD

Initial kinetic energy = 106.58 J. It means it had a velocity of

0.5mv² = 106.58

v = 7.3 m/s , momentum = 4*7.3 = 29.2 kgm/s

Now when its kinetic energy was multiplied by 1.78, it became 189.7124 J.

Then the velocity became v = 9.739 m/s, momentum = 4*9.739 = 38.956 kgm/s.

Therefore, momentum was multiplied by a factor of 1.334

SHORT METHOD

We know that Kinetic energy is directly proportional to square of velocity, while momentum is directly proportional to velocity. Therefore, if kinetic energy is increased by 1.78 . Then momentum is increased by sqrt(1.78) = 1.334

2) Now, as the foce is applied , the object will accelerate.

a = delta v/ delta t

a = (9.739 - 7.3) / 2.7 = 0.90333 m/s²

Thereofore, F = ma gives F = 4*0.90333 = 3.61332 N.

c) If the force is applied to the perpendicular direction , the speed will remain the same however the direction will change.