Question

Read the following and answer the given questions:
An object of mass 10 Kg is dropped from a height of 5 m. the different energies of the object at different heights are given in the following table:
Height at which the object is placed(m)
Kinetic energy(J)
Potential energy(J)
Total energy KE+PE(J)

5
0
500
500 J

4
100
400
500 J

3
200
300
500 J

2
300
200
500 J

1
400
100
500 J

JUST ABOVE THE GROUND
500
0
500 J



Q (i) Explain the law on which the given table is based. 2
Q (ii) What is the work to be done to increase the velocity of a car from 35 Km/h to 70 Km/ h if the mass of the car is 1200 Kg? ​

Answer 1

Answer:

Mass of bullet, m=100g=

1000

100

kg=

10

1

kg

Velocity of bullet, v=100ms

−1

Mass of gun, M=20kg

Let recoil velocity of gun =V

Before firing, the system (gun+bullet) is at rest, therefore, initial momentum of the system =0

Final momentum of the system = momentum of bullet + momentum of gun

=mv+MV=

10

1

×100+20V=10+20V

Applying law of conservation of momentum, Final momentum = Initial momentum

i.e., 10+20V=0

20V=−10

or V=−

20

10

=−0.5ms

−1

.

Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.