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Physics NEET Questions — Free MCQ Practice

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Which of the following is not a perfectly inelastic collision

A.

Striking of two glass balls

B.

A bullet striking a bag of sand

C.

An electron captured by a proton

D.

A man jumping onto a moving cart

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Explanation

Because in perfectly inelastic collision the colliding bodies stick together and move with common velocity

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A neutron having mass of $1 .67 \times 10^{- 27} k g$ and moving at $10^{8} m / s$ collides with a deutron at rest and sticks to it. If the mass of the deutron is $3 .34 \times 10^{- 27} k g$ then the speed of the combination is

A.

$2 .56 \times 10^{3} m / s$

B.

$2 .98 \times 10^{5} m / s$

C.

$3 .33 \times 10^{7} m / s$

D.

$5 .01 \times 10^{9} m / s$

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Explanation

According to law of conservation of momentum.

Momentum of neutron = Momentum of combination

⇒ $1 .67 \times 10^{- 27} \times 10^{8} = (1 .67 \times 10^{- 27} + 3 .34 \times 10^{- 27}) v$

∴ $v = 3 .33 \times 10^{7} m / s$

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A body of mass m₁ is moving with a velocity V. It collides with another stationary body of mass m₂. They get embedded. At the point of collision, the velocity of the system

A.

Increases

B.

Decreases but does not become zero

C.

Remains same

D.

Become zero

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Explanation

By momentum conservation before and after collision.

$m_{1} V + m_{2} \times 0 = (m_{1} + m_{2}) v$ ⇒ $v = \frac{m_{1}}{m_{1} + m_{2}} V$

i.e. Velocity of system is less than V.

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A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be

A.

$\frac{1}{2} m v^{2} \times \frac{m}{(m + M)}$

B.

$\frac{1}{2} m v^{2} \times \frac{M}{(m + M)}$

C.

$\frac{1}{2} m v^{2} \times \frac{(M + m)}{M}$

D.

$\frac{1}{2} M v^{2} \times \frac{m}{(m + M)}$

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Explanation

By conservation of momentum, $m v + M \times 0 = (m + M) V$

Velocity of composite block $V = (\frac{m}{m + M}) v$

K.E. of composite block $= \frac{1}{2} (M + m) V^{2}$

$= \frac{1}{2} (M + m) {(\frac{m}{M + m})}^{2} v^{2} = \frac{1}{2} m v^{2} (\frac{m}{m + M})$

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A shell is fired from a cannon with velocity v m/sec at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed (in m/sec) of the other piece immediately after the explosion is

A.

3v cos θ

B.

2v cos θ

C.

$\frac{3}{2} v \cos θ$

D.

$\frac{\sqrt{3}}{2} v \cos θ$

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Two particles of masses m₁ and m₂ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time t = 0. They collide at time t₀. Their velocities become ${\vec{v}}_{1}'$ and ${\vec{v}}_{2}'$ at time 2t₀ while still moving in air. The value of $|(m_{1} \vec{v_{1}}' + m_{2} \vec{v_{2}}') - (m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}})|$ is

A.

Zero

B.

$(m_{1} + m_{2}) g t_{0}$

C.

$2 (m_{1} + m_{2}) g t_{0}$

D.

$\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

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Explanation

The momentum of the two-particle system, at t = 0 is

${\vec{P}}_{i} = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}$

Collision between the two does not affect the total momentum of the system.

A constant external force $(m_{1} + m_{2}) g$ acts on the system.

The impulse given by this force, in time t = 0 to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$

∴ |Change in momentum in this interval|

$= |m_{1} \vec{v}'_{1} + m_{2} \vec{v}'_{2} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})| = 2 (m_{1} + m_{2}) g t_{0}$

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Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision the projectile and the struck particle move in directions making angles θ₁ and θ₂respectively with the initial direction of motion. The sum of the angles. θ₁ + θ₂, is

A.

45°

B.

90°

C.

135°

D.

180°

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Explanation

If the masses are equal and target is at rest and after collision both masses moves in different direction. Then angle between direction of velocity will be 90°, if collision is elastic.

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A rope is wound around a hollow cylinder of mass 3 kg and radius 40cm. What is the angular acceleration of the cylinder,if the rope is pulled with a force of 30 N?

A.

25 $m / s^{2}$

B.

0.25 $r a d / s^{2}$

C.

25 $r a d / s^{2}$

D.

5 $m / s^{2}$

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Explanation

The force applied on the rope provides a torque about the axis of the cylinder, given by F Ã— r = 30 Ã— 0.4 = 12 Nm. The moment of inertia of a hollow cylinder about its axis is MR^2/2. Using torque = I Ã— Î±, we get Î± = 12/(3Ã—0.4^2/2) = 25 rad/s^2.

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Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $ω_{1} a n d ω_{2}$ . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this proces is

A.

$\frac{1}{2} I {(ω_{1} + ω_{2})}^{2}$

B.

$\frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

C.

$I {(ω_{1} - ω_{2})}^{2}$

D.

$\frac{1}{8} I {(ω_{1} - ω_{2})}^{2}$

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Explanation

(d) Thinking Process

When no external torque acts on system then, angular momentum of system remains constant.

Angular momentum before contact = $I_{1} ω_{1} + I_{2} ω_{2}$

Angular momentum after the discs brought into contact.

$= I_{n e t} ω = (I_{1} + I_{2}) ω$

So, final angular speed of system= $ω$

$= \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}}$

Now, to calculate loss of energy, we subtract initial and final energies of system. $\Rightarrow L o s s o f e n e r g y = \frac{1}{2} I ω_{1}^{2} + \frac{1}{2} I ω_{2}^{2} - \frac{1}{2} (2 I) ω^{2} = \frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

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A bullet of mass 10g moving horizontal with a velocity of 400 m/s strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges of horizontally from the block wiil be

A.

100 m/s

B.

80 m/s

C.

120 m/s

D.

160 m/s

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Explanation

(c) According to the law of consevation of momentum.

$p_{i} = p_{f} \Rightarrow (0.01) \times 400 + 0 = 2 v + (0.01) v^{'} . . (i)$

Also velocity v of the block just after the collision is

$v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} . . . (i i)$

$\Rightarrow F r o m E q s . (i) a n d (i i), w e h a v e v^{'} = 120 m / s$

Physics MCQs for NEET — Practice Questions with Answers

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