Two bodies of masses 2 m and m have their KE in the ratio 8 : 1. What is the ratio of their momenta ?

8 : 1
4 : 1
2 : 1
1 : 1

Solution

\(\frac{p_{1}}{p_{2}}=\sqrt{\frac{2m_{1}k_{1}}{2m_{2}k_{2}}}\)

Which one of the following physical quantities is represented by the shaded area in the given graph?

Torque
Impulse
Power
Work done

Solution

\(Work \: done=\int Fdx\)

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

18 J
21 J
26 J
13 J

Solution

Work done = area under F-d graph

\(=\left [ 2\times (7-2) \right ]+\left [ \frac{1}{2}\times 2(12-7) \right ]\)

= 8 + 5= 13 J

The potential energy of a system increases if work is done

upon the system by anon conservative force
by the system against a conservative force
by the system against anonconservative force
upon the system by a conservative force

Solution

When work is done upon a system by a conservative force then its potential energy increases.

A force F acting on an object varies with distance x as shown here. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 6 m is:

18.0 J
13.5 J
9.0 J
4.5 J

Solution

Work done = area under F-x graph

= area of trapezium OABC =¹⁄₂(3+6)(3)= 13.5 J

The temperature at the bottom of a high water fall is higher than that at the top because

by itself heat flows from higher to lower temperature
the difference in height causes a difference in pressure
thermal energy is transformed into mechanical energy
mechanical energy is transformed into thermal energy.

Solution

By the principle of conservation energy

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle,the motion of the particles takes place in a plane. It follows that

its kinetic energy is constant
its acceleration is constant
its velocity is constant
it moves in a straight line

Solution

Work done by such force is always zero since force is acting in a direction perpendicular to velocity
∴from work energy theorem=ΔK=0
K remains constant.

A mass m₁ moves with a great velocity. It strikes another mass m₂ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

m₁> m₂
m₁< m₂
m₁= m₂
cannot say

Solution

\(V_{1}=\frac{(m_{1}-m_{2})u_{1}}{m_{1}+m_{2}}\)As v₁ is negative and less than u₁, therefore, m₁< m₂

A body moves a distance of 10 m along a straight line under the action of a force of 5 newtons. If the work done is 25 joules, the angle which the force makes with the direction of motion of body is

0°
60°
30°
90°

Solution

When the kinetic energy of a body is increased to three times, then the momentum increases

6 times
1.732 times
\(\sqrt{2}\) times
2 times

Solution

If P = momentum, K = kinetic energy, then

\(p_{1}^{2}=2mk_{1},p_{2}^{2}=2mk_{2}\)

\(∴\left ( \frac{p_{2}}{p_{1}} \right )=\frac{k_{2}}{k_{1}}=\frac{3k_{1}}{k_{1}}=3∴\frac{p_{2}}{p_{1}}=\sqrt{\frac{3}{1}}=1.732\)

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