Two blocks of masses 2 kg and M are at rest on an inclined plane and are separated by a distance of

Q: Why can we conserve momentum during the collision if there is friction and gravity acting on the blocks?

During the extremely short duration of the collision ($\Delta t \to 0$), the impulsive normal forces between the blocks are infinitely larger than the finite non-impulsive forces like gravity and friction. Therefore, the impulse due to external forces is negligible, and momentum is conserved along the line of impact.

Q: Why do we use the Work-Energy Theorem instead of Newton's equations of motion ($v^2 = u^2 + 2as$)?

You absolutely can use kinematics! Since the forces (gravity and kinetic friction) are constant, the acceleration is constant. However, the Work-Energy Theorem ($\Delta KE = W_{\text{net}}$) is often faster and less prone to sign errors because it deals directly with scalars (energy and work) rather than vectors.

Q: In Stage 2, why is the work done by gravity positive?

Work is defined as the dot product of force and displacement ($W = \vec{F} \cdot \vec{s}$). When the block slides down the incline, the component of gravity along the incline ($mg \sin\theta$) points in the exact same direction as the displacement. Since the angle between them is $0^\circ$, the work done is positive.

Q: How did we approximate $\cos\theta$ from $\sin\theta \approx \tan\theta = 0.05$?

For very small angles, $\sin\theta \approx \tan\theta \approx \theta$ (in radians). We can find $\cos\theta$ using the Pythagorean identity: $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (0.05)^2} = \sqrt{1 - 0.0025} = \sqrt{0.9975}$. This is approximately $0.9987$, which we rounded to $0.99$ to simplify the calculations as per standard JEE approximations.

Q: Why is the separation velocity $v_2 + v_3$ and not $v_3 - v_2$?

Velocity of separation is the relative speed at which the two objects move apart. After the collision, the $2\text{ kg}$ block moves *down* the incline with speed $v_2$, while block $M$ moves *up* the incline with speed $v_3$. Since they are moving in opposite directions, their relative speed of separation is the sum of their individual speeds, $v_2 + v_3$.

Question

Two blocks of masses $2	ext{ kg}$ and $M$ are at rest on an inclined plane and are separated by a distance of $6.0	ext{ m}$ as shown. The coefficient of friction between each block and the inclined plane is $0.25$. The $2	ext{ kg}$ block is given a velocity of $10.0	ext{ m/s}$ up the inclined plane. It collides with $M$, comes back and has a velocity of $1.0	ext{ m/s}$ when it reaches its initial position. The other block $M$ after the collision moves $0.5	ext{ m}$ up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block $M$.
[Take $\sin 	heta \approx 	an 	heta = 0.05$ and $g = 10	ext{ m/s}^2$]

Sigma Percentile · Accepted Answer

### The Symphony of Mechanics: Work, Energy, and Collisions

Physics problems that weave together multiple core concepts are often the most rewarding to solve. They require us to shift our perspective, seamlessly transitioning from the macroscopic world of energy conservation to the microscopic, split-second realm of impulsive collisions. This problem is a perfect example of such a symphony. We have a $2	ext{ kg}$ block launched up a rough incline, a dramatic mid-air (or rather, mid-incline) collision with an unknown mass $M$, and the subsequent aftermath where both blocks slide to a halt.

To unravel this, we must dissect the motion into three distinct, non-collision phases where the Work-Energy Theorem reigns supreme, and one instantaneous collision phase governed by the Conservation of Linear Momentum.

### Setting the Stage: The Geometry of the Incline

Before we analyze the forces, we must establish our geometric foundation. We are given that $\sin	heta \approx 	an	heta = 0.05$. Because the block slides on a rough surface, kinetic friction will play a major role, and friction depends on the normal force, which in turn depends on $\cos	heta$.

Using the fundamental Pythagorean identity, we can find $\cos	heta$:

$$ \cos	heta = \sqrt{1 - \sin^2	heta} = \sqrt{1 - (0.05)^2} $$

$$ \cos	heta = \sqrt{1 - 0.0025} = \sqrt{0.9975} \approx 0.99 $$

With $\cos	heta \approx 0.99$ and $\sin	heta = 0.05$ locked in, we are ready to tackle the dynamics.

### Phase 1: The Ascent of the $2	ext{ kg}$ Block

The $2	ext{ kg}$ block is launched with an initial velocity of $10	ext{ m/s}$ and travels $6.0	ext{ m}$ up the incline before striking block $M$. During this ascent, both gravity and kinetic friction oppose the motion, doing negative work and draining the block's kinetic energy.

Let $v_1$ be the velocity of the $2	ext{ kg}$ block just a fraction of a millisecond before impact. We apply the Work-Energy Theorem ($\Delta KE = W_{	ext{net}}$):

$$ \frac{1}{2}m(v_1^2 - 10^2) = W_{	ext{friction}} + W_{	ext{gravity}} $$

$$ \frac{1}{2}m(v_1^2 - 100) = -\mu (mg\cos	heta)s - mg(s\sin	heta) $$

Notice the elegance of physics here: the mass $m$ cancels out entirely from the equation. Multiplying through by $2$, we get:

$$ v_1^2 - 100 = -2gs(\mu\cos	heta + \sin	heta) $$

Substituting our known values ($s = 6	ext{ m}$, $\mu = 0.25$, $g = 10	ext{ m/s}^2$):

$$ v_1^2 = 100 - 2(10)(6)(0.25 	imes 0.99 + 0.05) $$

$$ v_1^2 = 100 - 120(0.2475 + 0.05) = 100 - 120(0.2975) $$

$$ v_1^2 = 100 - 35.7 = 64.3 \implies v_1 \approx 8	ext{ m/s} $$

The block strikes $M$ with a speed of $8	ext{ m/s}$.

### Phase 2: The Rebound of the $2	ext{ kg}$ Block

Immediately after the collision, the $2	ext{ kg}$ block rebounds with a downward velocity $v_2$. It slides $6.0	ext{ m}$ back to its starting point, arriving with a residual speed of $1.0	ext{ m/s}$.

We apply the Work-Energy Theorem again, but we must be careful with our signs. Friction still opposes the motion (doing negative work), but gravity is now pulling the block *down* the incline, doing positive work.

$$ \frac{1}{2}m(1^2 - v_2^2) = -\mu (mg\cos	heta)s + mg(s\sin	heta) $$

$$ 1 - v_2^2 = 2gs(-\mu\cos	heta + \sin	heta) $$

Substituting the values:

$$ 1 - v_2^2 = 2(10)(6)(-0.25 	imes 0.99 + 0.05) $$

$$ 1 - v_2^2 = 120(-0.2475 + 0.05) = 120(-0.1975) $$

$$ 1 - v_2^2 = -23.7 \implies v_2^2 = 24.7 $$

Since $24.7$ is incredibly close to $25$, we can safely approximate $v_2 \approx 5	ext{ m/s}$. The block rebounds downwards at $5	ext{ m/s}$.

### Phase 3: The Upward Kick of Block $M$

The collision transfers momentum to block $M$, giving it an initial upward velocity $v_3$. It slides a short distance of $s' = 0.5	ext{ m}$ before friction and gravity bring it to a halt.

Applying the Work-Energy Theorem for block $M$:

$$ 0 - \frac{1}{2}Mv_3^2 = -\mu (Mg\cos	heta)s' - Mg(s'\sin	heta) $$

Again, the mass $M$ cancels out, and the negative signs vanish:

$$ v_3^2 = 2gs'(\mu\cos	heta + \sin	heta) $$

$$ v_3^2 = 2(10)(0.5)(0.25 	imes 0.99 + 0.05) $$

$$ v_3^2 = 10(0.2975) = 2.975 $$

Taking the square root, we find $v_3 = \sqrt{2.975} \approx 1.72	ext{ m/s}$.

### The Collision: Unveiling the Coefficient of Restitution

We now possess a complete kinematic snapshot of the moments immediately before and after the collision. The coefficient of restitution, $e$, is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact.

Before the collision, the $2	ext{ kg}$ block approaches the stationary block $M$ at $8	ext{ m/s}$. Thus, $v_{	ext{approach}} = 8	ext{ m/s}$.

After the collision, the $2	ext{ kg}$ block moves down the incline at $5	ext{ m/s}$, while block $M$ moves up the incline at $1.72	ext{ m/s}$. Because they are moving in opposite directions, they are separating at the sum of their speeds:

$$ v_{	ext{separation}} = v_2 + v_3 = 5 + 1.72 = 6.72	ext{ m/s} $$

Therefore, the coefficient of restitution is:

$$ e = \frac{v_{	ext{separation}}}{v_{	ext{approach}}} = \frac{6.72}{8} = 0.84 $$

### The Finale: Conservation of Momentum

To find the unknown mass $M$, we turn to the Conservation of Linear Momentum. During the infinitesimal duration of the collision, the impulsive normal forces between the blocks are astronomically larger than the finite forces of gravity and friction. Consequently, the external impulse is negligible, and momentum is conserved along the axis of the incline.

Taking 'up the incline' as the positive direction:

$$ 	ext{Initial Momentum} = m v_1 + M(0) = 2(8) = 16	ext{ kg m/s} $$

$$ 	ext{Final Momentum} = M v_3 - m v_2 = M(1.72) - 2(5) = 1.72M - 10 $$

Equating the initial and final momentum:

$$ 16 = 1.72M - 10 $$

$$ 1.72M = 26 $$

$$ M = \frac{26}{1.72} \approx 15.116	ext{ kg} $$

Rounding to two decimal places, we conclude that the mass of block $M$ is **$15.12	ext{ kg}$**.

By systematically breaking the problem into energetic phases and a momentum-conserving collision, we have successfully decoded the entire physical event.

Visualized Solution

Visualizing the Three Stages of Motion

Evaluating $cos θ$ from $sin θ$

Work-Energy Theorem: 2 kg Block Moving Up

Calculating $v_{1}$

Work-Energy Theorem: 2 kg Block Moving Down

Calculating $v_{2}$

Work-Energy Theorem: Block M Moving Up

Calculating $v_{3}$

Calculating the Coefficient of Restitution(e)

Conservation of Momentum Along the Incline

Calculating the Mass $M$

The Sigma Insight: Head-on Collision

The Symphony of Mechanics

Setting the Stage

Phase 1

Phase 2

Phase 3

The Collision

The Finale

Similar Questions

A block of mass $0.50 kg$ is moving with a speed of $2.00 ms^{- 1}$ on a smooth surface. It strikes another mass of $1.00 kg$ and then they move together as a single body. The energy loss during the collision is

A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body?

An object of mass $m_{1}$ collides with another object of mass $m_{2}$ , which is at rest. After the collision, the objects move with equal speeds in opposite direction. The ratio of the masses $m_{2} : m_{1}$ is