NEET Laws of Motion MCQs & PYQs

Practice NEET Laws of Motion Questions

Question 81:

moderate

A particle moves on a rough horizontal ground with some initial velocity say \(v_0\). If \(3/4^{\text{th}}\) of its kinetic energy is lost in friction in time \(t_0\). Then coefficient of friction between the particle and the ground is:

Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).

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Question 82:

moderate

A body of mass 1 kg has velocity \(1\text{ ms}^{-1}\), up an inclined plane of angle of \(30^\circ\) to the horizontal. The friction coefficient is \(\frac{1}{\sqrt{3}}\). The distance the body travels before stopping is (\(g = 10\text{ m s}^{-2}\)):

Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).

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Question 83:

moderate

A uniform chain of length \( L \) is placed on a rough horizontal table with some part hanging below the table. If the length of the hanging part becomes \( \frac{2L}{5} \) then the chain starts sliding on the table.

The co-efficient of friction between the chain and the table is:

The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).

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Question 84:

easy

A projectile is thrown at a speed of \(100\text{ m/s}\) at an angle of \(37^\circ\) above the horizontal. At the highest point the projectile breaks into two parts of mass ratio \(1 : 3\), the smaller coming to rest. Find the speed of the second piece.

At the highest point, velocity is horizontal: \(v_x = u\cos(37^\circ) = 100 \times 0.8 = 80\text{ m/s}\). By conservation of linear momentum along the horizontal: \(m v_x = m_1 v_1 + m_2 v_2\). Here \(m_1 = m/4\) (comes to rest, \(v_1 = 0\)) and \(m_2 = 3m/4\). Thus, \(m(80) = \frac{3m}{4} v_2\), which gives \(v_2 = \frac{320}{3}\text{ m/s}\).

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Question 85:

easy

A large force \(f\) acts on a particle of mass \(m\) for a short time \(t\). The impulse imparted to the particle is given by

Impulse is defined as the product of the average force and the short time interval over which it acts, i.e., \(\text{Impulse} = f \times t\).

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Question 86:

easy

A block of mass \(20\text{ kg}\) is placed on a rough horizontal surface, and it is acted upon by a horizontal force of \(40\text{ N}\). If the coefficient of friction is \(0.2\), then the acceleration of the block is

The maximum limiting frictional force is \(f_{max} = \mu m g = 0.2 \times 20 \times 10 = 40\text{ N}\) (using \(g = 10\text{ m/s}^2\)). Since the applied force of \(40\text{ N}\) is equal to the limiting friction, the net horizontal force on the block is zero, resulting in zero acceleration.

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Question 87:

easy

A ball of mass \(0.15\text{ kg}\) is dropped from a height \(10\text{ m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (\(g = 10\text{ m/s}^2\)) nearly

The velocity of the ball just before striking the ground is \(u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 14.14\text{ m/s}\). Since it rebounds to the same height, its velocity just after is \(v = 14.14\text{ m/s}\) upwards. The change in momentum is \(Delta p = m(v - (-u)) = 2mu = 2 \times 0.15 \times 14.14 \approx 4.24\text{ kg m/s}\).

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Question 88:

easy

An object of mass 2 kg is placed on a smooth horizontal surface. A water jet throws water at a rate of 2 kg/s with speed 20 m/s which strikes the object horizontally. The force applied by water jet and acceleration of the object respectively, are

Formula: \(F = v \frac{dm}{dt} = 20 \times 2 = 40\text{ N}\). Using Newton's second law, acceleration \(a = \frac{F}{m} = \frac{40}{2} = 20\text{ m/s}^2\).

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Question 89:

easy

A player catches a ball of mass \(150\text{ g}\) in \(0.1\text{ s}\) moving with speed \(20\text{ m/s}\), then he experiences an average force of

By Newton's second law, \(F = \frac{\Delta p}{\Delta t} = \frac{mv}{t}\). Substituting the values, \(F = \frac{0.15\text{ kg} \times 20\text{ m/s}}{0.1\text{ s}} = 30\text{ N}\).

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Question 90:

easy

A body is said to be in mechanical equilibrium if

For complete mechanical equilibrium, a body must be in both translational equilibrium (net external force is zero) and rotational equilibrium (net external torque is zero).

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