Practice Test 5 Archives

A triangular block of mass M with angles 30°, 60°, and 90°rests with its 30°–90° side on a horizontal table. A cubical block of mass m rests on the 60°–30° side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming friction less contact is

g
^g⁄_√2
^g⁄_√3
^g⁄_√5

Solution

ma cos 30° = mg sin 30°

∴ a = ^g⁄_√3

An object at rest in space suddenly explodes into three parts of same mass. The momentum of the two parts are \(2p\hat{i}\) and \(2p\hat{i}\). The momentum of the third part

will have a magnitude p√3
will have a magnitude p√5
will have a magnitude p
will have a magnitude 2p.

Solution

A body of mass 32 kg is suspended by a spring balance from the roof of a vertically operating lift and going downward from rest.At the instant the lift has covered 20 m and 50 m, the spring balance showed 30 kg and 36 kg respectively. Then the velocity of the lift is

decreasing at 20 m, and increasing at 50 m
increasing at 20m and decreasing at 50 m
continuously decreasing at a steady rate throughout the journey
constantly increasing at constant rate throughout the journey.

Solution

While moving down, when the lift is accelerating the weight will be less and when the lift is decelerating the weight will be more.

A 1 kg block and a 0.5 kg block move together on a horizontal friction less surface . Each block exerts a force of 6 N on the other. The block move with a uniform acceleration of

3 ms^-2
6 ms^-2
9 ms^-2
12 ms^-2

Solution

For 0.5 kg block, 6 = 0.5 a

For the arrangement shown in the Figure the tension in the string is [Given : tan^-1 (0.8) 39°]

6 N
6.4 N
0.4 N
zero

Solution

Here tan θ = 0.8

where θ is angle of repose

θ = tan^-1 (0.8) = 39°

The given angle of inclination is equal to the angle of repose. So the 1kg block has no tendency to move.

∴ mg sin θ= force of friction

⇒T =0

A bag of sand of mass m is suspended by a rope. A bullet of mass ^m⁄₂₀ is fired at it with a velocity v and gets embedded into it. The velocity of the bag finally is

^v⁄₂₀ × 21
^20v⁄₂₁
^v⁄₂₀
^v⁄₂₁

Solution

Applying law of conservation of momentum Momentum of bullet = Momentum of sand-bullet system

^m⁄₂₀ v=(m + ^m⁄₂₀) v=²¹⁄₂₀ mv

A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. If the force applied on the body is 2.5 N, then the frictional force acting on the body will be [g = 10 ms^-2]

8 N
10 N
20 N
2.5 N

Solution

Limiting friction= 0.5 × 2 × 10 = 10 N

The applied force is less than force of friction, therefore the force of friction is equal to the applied force.

A bird is in a wire cage which is hanging from a spring balance . In the first case, the bird sits in the cage and in the second case, the bird flies about inside the cage. The reading in the spring balance is

more in the first case
less in first case
unchanged
zero in second case.

Solution

Based on Newton’s third law of motion.

A block of mass 4 kg rests on an inclined plane.The inclination to the plane is gradually increased. It is found that when the inclination is 3 in 5, the block just begins to slide down the plane. The coefficient of friction between the block and the plane is

0.4
0.6
0.8
0.75

Solution

A rifle man, who together with his rifle has a mass of 100 kg,stands on a smooth surface and fires 10 shots horizontally.Each bullet has a mass 10 g and a muzzle velocity of 800 ms^-1. The velocity which the rifle man attains after firing 10 shots is

8 m s^-1
0.8 m s^-1
0.08 m s^-1
– 0.8 m s^-1

Solution

According to law of conservation of momentum,

100 v=- ¹⁰⁄₁₀₀₀ × 10 × 800

ie, v= 0.8 ms^-1