The potential energy of particle in a force field is U=^A⁄_r²–^B⁄_r, where A and Bare positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

B/ 2A
2A/ B
A/ B
B/ A

Solution

A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

\(\frac{1}{4}V\hat{i}+\frac{3}{2}V\hat{j}\)
\(\frac{1}{3}V\hat{i}+\frac{2}{3}V\hat{j}\)
\(\frac{2}{3}V\hat{i}+\frac{1}{3}V\hat{j}\)
\(\frac{3}{2}V\hat{i}+\frac{1}{4}V\hat{j}\)

Solution

A body projected vertically from the earth reaches a height equal to earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest

at the highest position of the body
at the instant just before the body hits the earth
it remains constant all through
at the instant just after the body is projected

Solution

Power exerted by a force is given by P =F.v When the body is just above the earth’s surface, its velocity is greatest. At this instant, gravitational force is also maximum. Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be

0, 1
1, 1
1, 0.5
0, 2

Solution

A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k= 50 N/m. The maximum compression of the spring would be

0.5 m
0.15 m
0.12 m
1.5 m

Solution

¹⁄₂ mv² =¹⁄₂ kx² or 0.5 × (1.5)² = 50 × x²

∴ x = 0.15 m

A stationary particle explodes into two particles of masses m₁ and m₂ which move in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies E₁/E₂ is

m₁v₂/m₂v₁
m₂/m₁
m₁/m₂
1

Solution

In a simple pendulum of length ι the bob is pulled a side from its equilibrium position through an angle θ and then released. The bob passes through the equilibrium position with speed

\(\sqrt{2g\iota (1+cos\theta )}\)
\(\sqrt{2g\iota\, sin\, \theta}\)
\(\sqrt{2g\iota}\)
\(\sqrt{2g\iota(1-cos\, \theta )}\)

Solution

A bomb of mass 1 kg is thrown vertically upwards with a speed of 100 m/s. After 5 seconds it explodes into two fragments.One fragment of mass 400 gm is found to go down with a speed of 25 m/s. What will happen to the second fragment just after the explosion? (g = 10 m/s²)

It will go upward with speed 40 m/s
It will go upward with speed 100 m/s
It will go upward with speed 60 m/s
It will also go downward with speed 40m/s

Solution

A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/s at an angle of 60º.It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for 0.20 s,what is the average force exerted on the ball by the wall?

150 N
Zero
150 √3N
300 N

Solution

A rubber ball is dropped from a height of 5m on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of

¹⁶⁄₂₅
²⁄₅
³⁄₅
⁹⁄₂₅

Solution

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