A ball of mass 0.5 kg moving with a velocity of 2m/sec strikes a wall normally and bounces back with the same speed.If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is :

2000 newton
1000 newton
5000 newton
125 newton

Solution

\(F=\frac{mv-(-mv)}{t}=\frac{2mv}{t}=\frac{2\times 0.5\times 2}{10^{-3}}=2\times 10^{3}N\)

Starting from rest, a body slides down a 45º inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

0.33
0.25
0.75
0.80

Solution

A man weighing 80 kg is standing on a trolley weighing 320kg. The trolley is resting on friction less horizontal rails. If the man starts walking on the trolley along the rails at a speed of one metre per second, then after 4 seconds, his displacement relative to the ground will be :

5 metres
4.8 metres
3.2 metres
3.0 metres

Solution

Displacement of the man on the trolley = 1 × 4 = 4m

Now applying conservation of linear momentum

80 × 1 + 400 v = 0 or V=- ¹⁄₅ m/sec.

The distance travelled by the trolley

= – 0.2 × 4 = –0.8 m.

(In opposite direction to the man.)

Thus, the relative displacement of the man with the ground = (4 – 0.8) = 3.2 m.

A spring is compressed between two toy carts of mass m₁ and m₂. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v₁ and v₂ are the velocities of the toy carts and there is no friction between the toy carts and the ground,then :

v₁/v₂= m₁/m₂
v₁/v₂= m₂/m₁
v₁/v₂ = –m₂/m₁
v₁/v₂ = –m₁/m₂

Solution

Applying law of conservation of linear momentum

m₁v₁+ m₂v₂= 0,^m₁⁄_m₂=-^V₂⁄_V₁

A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving northward with the same speed v. The two particles coalesce on collision. The new particle of mass 2m will move in the north-external direction with a velocity :

v/2
2v
v /√2
None of these

Solution

A rocket has a mass of 100 kg. Ninety percent of this is fuel. It ejects fuel vapors at the rate of 1 kg/sec with a velocity of 500m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial upthrust on the rocket when it just starts moving upwards is

zero
500 newton
1000 newton
2000 newton

Solution

A cart of mass M has a block of mass m attached to it as shown in fig. The coefficient of friction between the block and the cart is μ. What is the minimum acceleration of the cart so that the block m does not fall?

μg
g/μ
μ/g
M μg/m

Solution

See fig.

If a= acceleration of the cart, then N = ma

∴ μN = mg or μ ma= mg or a = g/μ

Two blocks of masses 2 kg and 1 kg are placed on a smooth horizontal table in contact with each other. Ahorizontal force of 3 newton is applied on the first so that the block moves with a constant acceleration. The force between the blocks would be

3 newton
2 newton
1 newton
zero

Solution

See fig. Let F be the force between the blocks and a their common acceleration. Then for 2 kg block,

3 –F =2 a ...(1)

for 1kg block, F = 1 × a = a ....(2)

∴3 –F = 2 For 3 F= 3 or F =1 newton

The elevator shown in fig. is descending with an acceleration of 2 m/s². The mass of the block A= 0.5 kg. The force exerted by the block A on block B is

Solution

R= mg –ma = 0.5 × 10 –0.5 ×2 = 5 –1 = 4 N

Consider the system shown in fig. The pulley and the string are light and all the surfaces are friction less. The tension in the string is (take g= 10 m/s²)

Solution

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