JEE Mains Laws Of Motion: Previous Year Questions & Detailed Solutions
This post contains previous year questions on the Laws Of Motion chapter from JEE Mains along with detailed solutions. Choose an option to check your answer and click "Show Detailed Solution" to see the step-by-step explanation.
A monkey of mass 50 kg climbs on a rope which can withstand the tension (T) of 350 N. If the monkey initially climbs down with an acceleration of \(4\,m/s^2\) and then climbs up with an acceleration of \(5\,m/s^2\) (take \(g = 10\,m/s^2\)), choose the correct option:
Detailed Solution:
Climbing Down: When the monkey climbs down with acceleration \(a = 4\,m/s^2\), the net force acting upward is given by:
\(\,T = mg - ma = 50 \times 10 - 50 \times 4 = 500 - 200 = 300\,N.\)
Climbing Up: When climbing up with \(a = 5\,m/s^2\), the net force equation becomes:
\(\,T = mg + ma = 50 \times 10 + 50 \times 5 = 500 + 250 = 750\,N.\)
Since the rope can withstand only \(350\,N\), it will break when the monkey climbs upward. Hence, the correct option is C.
A bag is gently dropped on a conveyor belt moving at a speed of \(2\,m/s\). The coefficient of friction between the belt and the bag is 0.4. Initially, the bag slips on the belt before coming to rest due to friction. The distance travelled by the bag on the belt during the slipping motion is:
Detailed Solution:
The frictional acceleration is given by: \(a = \mu g = 0.4 \times 10 = 4\,m/s^2\).
The bag initially has a relative speed of \(2\,m/s\) (with respect to the belt).
The distance covered during slipping is found using:
\[
s = \frac{v^2}{2a} = \frac{2^2}{2 \times 4} = \frac{4}{8} = 0.5\,m.
\]
Thus, the correct answer is B.
A block of mass \(M\) slides down on a rough inclined plane with constant velocity. The angle made by the incline with the horizontal is \(\theta\). The magnitude of the contact force (the resultant of normal reaction and friction) is:
Detailed Solution:
Since the block moves with constant velocity, the net force along the incline is zero. The frictional force balances the downslope component \(mg\sin\theta\) and the normal reaction is \(mg\cos\theta\). The resultant contact force is: \[ F_{\text{contact}} = \sqrt{(mg\cos\theta)^2 + (mg\sin\theta)^2} = mg\,\sqrt{\cos^2\theta+\sin^2\theta} = mg. \]
Thus, the correct answer is A.
A block 'A' takes 2 seconds to slide down a frictionless incline of angle \(30^\circ\) and length \(I\) inside a lift moving upward with uniform velocity. If the incline is changed to \(45^\circ\), the time taken by the block to slide down will be approximately:
Detailed Solution:
For a frictionless incline, the acceleration along the plane is \(g\sin\theta\). Using the equation: \[ l = \frac{1}{2}g\sin\theta\,t^2, \] for \(\theta=30^\circ\) and \(t=2\,s\) we get: \[ l = \frac{1}{2}\times 10\times\frac{1}{2}\times 4 = 10\,m. \] For \(\theta=45^\circ\) (\(\sin45^\circ=\frac{\sqrt{2}}{2}\)): \[ t = \sqrt{\frac{2l}{g\sin45^\circ}} = \sqrt{\frac{20}{10\,\times\,\frac{\sqrt{2}}{2}}} \approx 1.68\,s. \]
Thus, the correct answer is C.
An object of mass \(5\,kg\) moving at \(25\,m/s\) hits two different walls and comes to rest in (i) 3 seconds and (ii) 5 seconds respectively. Choose the correct option:
Detailed Solution:
The impulse imparted to the object is its change in momentum: \[ \Delta p = m\,v = 5 \times 25 = 125\,Ns, \] which is the same in both cases. However, the average force is given by: \[ F_{\text{avg}} = \frac{\Delta p}{\Delta t}, \] so for 3 seconds, \(F_{\text{avg}} \approx \frac{125}{3} \approx 41.67\,N\) and for 5 seconds, \(F_{\text{avg}} = \frac{125}{5} = 25\,N\).
Thus, the correct answer is B.
A balloon of mass \(10\,g\) has air escaping at a uniform rate with a velocity of \(4.5\,cm/s\). If the balloon shrinks completely in 5 seconds, then the average force acting on the balloon is (in dyne):
Detailed Solution:
The rate of mass loss is \(\frac{10\,g}{5\,s} = 2\,g/s\). The force due to the escaping air is the rate of change of momentum:
\[
F = (\text{mass loss rate}) \times (\text{velocity}) = 2 \times 4.5 = 9\,dyne.
\]
(Recall that \(1\,g\cdot cm/s^2 = 1\,dyne\).)
Thus, the correct answer is B.
A boy pushes a box of mass \(2\,kg\) with a force \(\vec{F} = 20\,\hat{i} + 10\,\hat{j}\,N\) on a frictionless surface. If the box is initially at rest, the displacement along the X-axis after 10 seconds is:
Detailed Solution:
Only the \(x\)-component of the force contributes to the horizontal motion. Thus, \[ a_x = \frac{20}{2} = 10\,m/s^2. \] The displacement is given by: \[ s_x = \frac{1}{2} a_x t^2 = \frac{1}{2} \times 10 \times 10^2 = 500\,m. \]
Therefore, the correct answer is A.
The coefficient of static friction between a wooden block of mass \(0.5\,kg\) and a vertical rough wall is 0.2. The magnitude of the horizontal force that must be applied to keep the block from falling is:
Detailed Solution:
To prevent the block from falling, the frictional force must balance the weight, i.e., \[ \mu F = mg \quad\Longrightarrow\quad F = \frac{mg}{\mu} = \frac{0.5 \times 10}{0.2} = 25\,N. \]
Thus, the correct answer is B.
An inclined plane is shaped so that its vertical cross‑section is given by \(y = \frac{x^2}{4}\), where \(y\) is vertical and \(x\) is horizontal. If the surface is rough with a coefficient of friction \(\mu = 0.5\), the maximum height (in cm) at which a stationary block will not slip downward is:
Detailed Solution:
For a block on a curved surface, equilibrium occurs when the tangential component of gravity is balanced by the maximum static friction. At any point, the slope of the curve is \(\tan\theta = \frac{dy}{dx} = \frac{x}{2}\). Equilibrium requires: \[ \tan\theta \leq \mu \quad\Longrightarrow\quad \frac{x}{2} \leq 0.5 \quad\Longrightarrow\quad x \leq 1. \] The corresponding height is: \[ y = \frac{x^2}{4} = \frac{1^2}{4} = 0.25\,m = 25\,cm. \]
Thus, the correct answer is A.
A bullet of mass \(0.1\,kg\) is fired at a wooden block. It pierces the block and stops after traveling \(50\,cm\) into it. If the bullet's speed before impact is \(10\,m/s\) and it decelerates uniformly, then the magnitude of the effective retarding force (in N) is:
Detailed Solution:
Using the kinematic equation \(v^2 = u^2 + 2as\) with \(v=0\), \(u=10\,m/s\), and \(s=0.5\,m\): \[ 0 = 10^2 + 2a(0.5) \quad\Longrightarrow\quad a = -100\,m/s^2. \] The magnitude of the retarding force is: \[ F = m|a| = 0.1 \times 100 = 10\,N. \]
Thus, the correct answer is C.
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