Two blocks A and B of equal masses are released from an inclined plane of inclination 45° at t = 0.

Q: Wait, why do we use kinetic friction instead of static friction?

The problem states the blocks are "released", implying they start sliding. Since the angle is $45^\circ$ and $\tan 45^\circ = 1$, which is much greater than both $\mu_A (0.2)$ and $\mu_B (0.3)$, the downward component of gravity easily overcomes static friction. Thus, they are in motion, and we must use kinetic friction.

Q: Why can't we just use energy conservation?

While energy conservation is a powerful tool, it's not ideal here because friction is a non-conservative force that dissipates energy as heat. More importantly, we need to find the *time* it takes for them to meet, and energy equations don't explicitly contain time. Kinematics is the way to go!

Q: What if block B had a lower coefficient of friction than block A?

If $\mu_B < \mu_A$, block B would accelerate faster than block A. Since block B is already ahead, the distance between them would continuously increase, and they would never meet!

Q: Why is the relative initial velocity zero?

Both blocks are released from rest at $t = 0$. Since $u_A = 0$ and $u_B = 0$, their relative initial velocity $u_{\text{rel}} = u_A - u_B = 0 - 0 = 0$.

Q: Is it mandatory to use relative motion?

Not at all! You can solve it from the ground frame by setting up two equations: $s_A = \frac{1}{2} a_A t^2$ and $s_B = \frac{1}{2} a_B t^2$. Since A must cover the initial gap, $s_A = s_B + \sqrt{2}$. Substituting the equations will give you the exact same result, but relative motion just makes the algebra much cleaner!

Question

Two blocks $A$ and $B$ of equal masses are released from an inclined plane of inclination $45^{\circ}$ at $t = 0$. Both the blocks are initially at rest. The coefficient of kinetic friction between the block $A$ and the inclined plane is $0.2$ while it is $0.3$ for block $B$. Initially the block $A$ is $\sqrt{2}	ext{ m}$ behind the block $B$. When and where their front faces will come in a line?
(Take $g=10	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing next to a steep $45^\circ$ inclined plane. Two blocks, $A$ and $B$, are placed on this slope, with block $A$ trailing block $B$ by a precise distance of $\sqrt{2} 	ext{ m}$. Both are released from rest simultaneously.

At first glance, you might think they will just slide down together, maintaining that exact gap. But physics is in the details! The surface isn't uniformly smooth for both. Block $A$ experiences a kinetic friction coefficient of $\mu_A = 0.2$, while block $B$ faces a rougher patch with $\mu_B = 0.3$.

Because block $A$ faces less resistance, it will accelerate faster than block $B$. It's a race, and block $A$ is guaranteed to catch up. Our mission is to find out exactly *when* and *where* this collision happens.

### The Master Equation

To solve this, we first need to determine how fast each block is accelerating. Let's draw a free body diagram for a generic block of mass $m$ on an incline of angle $	heta$.

Gravity pulls the block straight down with a force of $mg$. We resolve this into two components: $mg \sin 	heta$ parallel to the incline (driving it down) and $mg \cos 	heta$ perpendicular to the incline (pressing it into the surface). The normal force $N$ perfectly balances the perpendicular component, so $N = mg \cos 	heta$.

Friction opposes the motion, acting upwards along the incline. Since the blocks are sliding, we use kinetic friction, given by $f_k = \mu N = \mu mg \cos 	heta$.

Using Newton's Second Law, the net force down the incline is:

$$F_{	ext{net}} = mg \sin 	heta - \mu mg \cos 	heta$$

Dividing by mass $m$, we get the master equation for acceleration:

$$a = g \sin 	heta - \mu g \cos 	heta$$

### Calculating the Accelerations

Now, let's arm ourselves with the given values: $g = 10 	ext{ m/s}^2$ and $	heta = 45^\circ$. Remember that $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$.

For block $A$, we substitute $\mu_A = 0.2$:

$$a_A = 10 \left(\frac{1}{\sqrt{2}}ight) - 0.2 	imes 10 \left(\frac{1}{\sqrt{2}}ight)$$

$$a_A = \frac{10}{\sqrt{2}} - \frac{2}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2} 	ext{ m/s}^2$$

For block $B$, we substitute $\mu_B = 0.3$:

$$a_B = 10 \left(\frac{1}{\sqrt{2}}ight) - 0.3 	imes 10 \left(\frac{1}{\sqrt{2}}ight)$$

$$a_B = \frac{10}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}} = 3.5\sqrt{2} 	ext{ m/s}^2$$

As predicted, $a_A > a_B$. Block $A$ is closing the gap!

### The Magic of Relative Kinematics

We could calculate the individual distances $s_A$ and $s_B$ from the ground frame and set $s_A = s_B + \sqrt{2}$. But why do the hard work when we can use a brilliant shortcut? Let's shift our perspective and "sit" on block $B$.

In block $B$'s frame of reference, block $B$ is completely stationary. Block $A$ is moving towards it with a relative acceleration:

$$a_{	ext{rel}} = a_A - a_B = 4\sqrt{2} - 3.5\sqrt{2} = 0.5\sqrt{2} = \frac{1}{\sqrt{2}} 	ext{ m/s}^2$$

Since both blocks started from rest, their initial relative velocity is zero ($u_{	ext{rel}} = 0$). The distance block $A$ needs to cover in this relative frame is simply the initial gap, $s_{	ext{rel}} = \sqrt{2} 	ext{ m}$.

### Final Calculation

We can now use the second equation of motion in the relative frame:

$$s_{	ext{rel}} = u_{	ext{rel}} t + \frac{1}{2} a_{	ext{rel}} t^2$$

Substituting our values:

$$\sqrt{2} = 0 + \frac{1}{2} \left(\frac{1}{\sqrt{2}}ight) t^2$$

Cross-multiplying to isolate $t^2$:

$$t^2 = \sqrt{2} 	imes 2 	imes \sqrt{2} = 4$$

Taking the square root, we find the time of impact:

$$t = 2 	ext{ s}$$

Finally, to find *where* they meet, we switch back to the ground frame and calculate the distance travelled by block $A$ in these 2 seconds:

$$s_A = \frac{1}{2} a_A t^2 = \frac{1}{2} (4\sqrt{2}) (2)^2$$

$$s_A = \frac{1}{2} 	imes 4\sqrt{2} 	imes 4 = 8\sqrt{2} 	ext{ m}$$

**Conclusion:** The blocks will align their front faces exactly **2 seconds** after release, at a distance of **$8\sqrt{2} 	ext{ m}$** down the inclined plane.

Visualized Solution

$Visualizing the Setup$

$Acceleration on an Incline$

$Acceleration of Block A$

$Calculating a_{A}$

$Acceleration of Block B$

$Calculating a_{B}$

$Relative Kinematics$

$Calculating a_{rel}$

$Time to Meet$

$Distance Travelled by A$

$Final Result$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation

Calculating the Accelerations

The Magic of Relative Kinematics

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration on an Incline

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration of Block A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating aA​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration of Block B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating aB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Kinematics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating arel​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time to Meet

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance Travelled by A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Result