Laws of Motion — Page 6 | NEET Physics

Q51

A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by

A 68%

B 41%

C 200%

D 100%

Correct Answer

Option B

Solution

When a stone is dropped from a height h, it hits the ground with a momentum

P = m\sqrt {2gh}

....(1) where m is the mass of the stone.

When the same stone is dropped from a height 2h (i.e. 100% of initial), then its momentum with which it hits the ground becomes

P' = m\sqrt {2g\left( {2h} \right)} = \sqrt 2 P

[Using (1)] ...(2) % change in momentum =

{{P' - P} \over P} \times 100\%

= {{\sqrt 2 P - P} \over P} \times 100\% = 41\%

Q52

A conveyor belt is moving at a constant speed of 2 ms

-

1 . A box is gently dropped on it. The coefficient of friction between them is

\mu

= 0.5. The distance that the box will move relativce to belt before coming to rest on it, taking g = 10 m s

-

2, , is

A 0.4 m

B 1.2 m

C 0.6 m

D zero

Correct Answer

Option A

Solution

Force of friction, f = $\mu$ mg $\therefore$

a = {f \over m} = {{\mu mg} \over m} = \mu g

= 0.5 \times 10 = 5

ms -2 Using v 2 – u 2 = 2aS 0 2 – 2 2 = 2(–5) $\times$ S $\Rightarrow$ S = 0.4 m

Q53

A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is

A MV

B 1.5 MV

C 2MV

D zero

Correct Answer

Option C

Solution

Impulse experienced by the body = change in momentum = MV – (–MV) = 2MV.

Q54

A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s 2 . If g = 10 ms

-

2 , the tension in the supporting cable is

A 8600 N

B 9680 N

C 11000 N

D 1200 N

Correct Answer

Option C

Solution

Here, Mass of a person, m = 60 kg Mass of lift, M = 940 kg, a = 1 m/s 2 , g = 10 m/s 2 Let T be the tension in the supporting cable.

$\therefore$ T – (M + m)g = (M + m)a $\Rightarrow$ T = (M + m)(a + g)= (940 + 60)(1 + 10) = 11000 N

Q55

A block of mass m is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is

\mu

. The acceleration

\alpha

of the cart that will prevent the block from falling satisfies

A

\alpha > {{mg} \over \mu }

B

\alpha > {g \over {\mu m}}

C

\alpha \ge {g \over \mu }

D

\alpha < {g \over \mu }

Correct Answer

Option C

Solution

Forces acting on the block are as shown in the fig.

Normal reaction N is provided by the force ma due to acceleration $\alpha$ $\therefore$ N = m $\alpha$ For the block not to fall, frictional force,

{F_f} \ge mg

\Rightarrow \mu N \ge mg

\Rightarrow \mu m\alpha \ge mg

\Rightarrow \alpha \ge g/\mu

Q56

A body, under the action of a force

\overrightarrow F = 6\widehat i - 8\widehat j + 10\widehat k,

acquires an accelerations of 1 m/s 2 . The mass of this body must be

A 10 kg

B 20 kg

C 10

\sqrt 2

kg

D 2

\sqrt {10}

kg

Correct Answer

Option C

Solution

\overrightarrow F = 6\widehat i - 8\widehat j + 10\widehat k

\left| {\overrightarrow F } \right| = \sqrt {36 + 64 + 100}

= \sqrt {200} N = 10\sqrt 2 N

Acceleration, a = 1 m s –2 $\therefore$

Mass,\,M = {{10\sqrt 2 } \over 1} = 10\sqrt 2 \,kg

Q57

The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is :

A 4 m s

-

2 upwards

B 4 m s

-

2 downwards

C 14 m s

-

2 upwards

D 30 m s

-

2 downwards

Correct Answer

Option A

Solution

F – Mg = Ma 8000 = 2000a $\therefore$ Acceleration is 4 ms –2 upwards.

Q58

Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

A

{{\sqrt 3 } \over 4}N

B

{\sqrt 3 }

N

C 0.5 N

D 1.5 N

Correct Answer

Option C

Solution

The components of 1N and 2N forces along + x axis = 1 cos 60° + 2 sin 30°

= 1 \times {1 \over 2} + 2 \times {1 \over 2} = {1 \over 2} + 1 = {3 \over 2} = 1.5N

The component of 4 N force along – x-axis

= 4\sin {30^o} = 4 \times {1 \over 2} = 2N

Therefore, if a force of 0.5N is applied along + x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axis.

Q59

A roller coaster is designed such that riders experience ''weightlessness'' as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between

A 16 m/s and 17 m/s

B 13 m/s and 14 m/s

C 14 m/s and 15 m/s

D 15 m/s and 16 m/s

Correct Answer

Option C

Solution

For the riders to experience weightlessness at the top of the hill, the weight of the rider must be balanced by the centripetal force. i.e.,

mg = {{m{v^2}} \over R}

\Rightarrow v = \sqrt {gR} = \sqrt {10 \times 20} = 14.1\,m{s^{ - 1}}

Hence, the speed of the car should be between 14 ms –1 and 15 ms –1 .

Q60

Sand is being dropped on a conveyer belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of

\upsilon

m/s will be :

A

{{Mv} \over 2}

newton

B zero

C M

v

newton

D 2 M

v

newton

Correct Answer

Option C

Solution

F = {d \over {dt}}\left( {Mv} \right) = v{{dM} \over {dt}} + M{{dv} \over {dt}}

As v is a constant,

F = v{{dM} \over {dt}}

But

{{dM} \over {dt}} = M\,kg/s

$\therefore$ To keep the conveyer belt moving at v m/s, force needed = vM newton.