Physics Q30 – JEE-MAIN 2026

Home Physics Laws of MotionQuestion #215

A wedge

Y

with mass of

10 kg

and all frictionless surfaces and the inclined surface making

37^{\circ}

with horizontal. A block

X

with mass

2 kg

is placed at the highest point of the wedge as shown in figure is at rest. At

t = 0

wedge

(Y)

is pulled toward right with constant force

(f)

24 N

. Taking the block

X

at rest at

t = 0

, the time taken by it to slide down

8.8 m

on the slope, while

Y

is on the move, is _______

s

. (take

\tan (37^{\circ}) = 3 / 4

and

g = 10 {m/s}^{2}

)

Clue 1 of 3: Frame of Reference

Analyze the motion of the block relative to the wedge by working in the non-inertial frame of the accelerating wedge.

🧠 Full Solution Path

Step 1: Analyze Forces and Accelerations in Wedge's Frame✦ Active

Given $m_{X} = 2 kg$ , $m_{Y} = 10 kg$ , $F = 24 N$ , $g = 10 {m/s}^{2}$ , $\tan (37^{\circ}) = 3 / 4$ . This implies $\sin (37^{\circ}) = 3 / 5$ and $\cos (37^{\circ}) = 4 / 5$ . Let $a_{Y}$ be the horizontal acceleration of the wedge to the right, and $a_{X / Y}$ be the acceleration of block $X$ relative to wedge $Y$ down the incline. In the non-inertial frame of the wedge, block $X$ experiences a pseudo force $m_{X} a_{Y}$ to the left. Applying Newton's second law to block $X$ along and perpendicular to the incline:

N_{X} - m_{X} g \cos θ - m_{X} a_{Y} \sin θ = 0 \Rightarrow N_{X} = m_{X} g \cos θ + m_{X} a_{Y} \sin θ (1)

m_{X} g \sin θ - m_{X} a_{Y} \cos θ = m_{X} a_{X / Y} \Rightarrow a_{X / Y} = g \sin θ - a_{Y} \cos θ (2)

💡 Teacher's Secret Hint

Remember to include the pseudo force when working in a non-inertial frame. The pseudo force acts opposite to the acceleration of the frame.

Step 2: Analyze Forces on the Wedge in Ground Frame○ Expand

Applying Newton's second law to wedge $Y$ in the horizontal direction (ground frame):

F - N_{X} \sin θ = m_{Y} a_{Y} (3)

Substitute $N_{X}$ from (1) into (3) to find $a_{Y}$ :

F - (m_{X} g \cos θ + m_{X} a_{Y} \sin θ) \sin θ = m_{Y} a_{Y}

F - m_{X} g \sin θ \cos θ - m_{X} a_{Y} \sin^{2} θ = m_{Y} a_{Y}

a_{Y} = \frac{F - m_{X} g \sin θ \cos θ}{m_{Y} + m_{X} \sin^{2} θ}

a_{Y} = \frac{24 - (2) (10) (3 / 5) (4 / 5)}{10 + (2) (3 / 5)^{2}} = \frac{24 - 9.6}{10 + 18 / 25} = \frac{14.4}{10 + 0.72} = \frac{14.4}{10.72} = \frac{90}{67} {m/s}^{2}

💡 Teacher's Secret Hint

Ensure correct signs for force components. The horizontal component of the normal force from the block on the wedge acts opposite to the applied force $F$ .

Step 3: Calculate Relative Acceleration and Time○ Expand

Now, substitute $a_{Y}$ into equation (2) to find $a_{X / Y}$ :

a_{X / Y} = g \sin θ - a_{Y} \cos θ = 10 (3 / 5) - (\frac{90}{67}) (4 / 5)

a_{X / Y} = 6 - \frac{72}{67} = \frac{402 - 72}{67} = \frac{330}{67} {m/s}^{2}

The block slides down a distance $d = 8.8 m$ relative to the wedge. Using the kinematic equation $d = v_{0} t + \frac{1}{2} a_{X / Y} t^{2}$ , with $v_{0} = 0$ :

d = \frac{1}{2} a_{X / Y} t^{2} \Rightarrow t = \sqrt{\frac{2 d}{a_{X / Y}}}

t = \sqrt{\frac{2 \times 8.8}{330 / 67}} = \sqrt{\frac{17.6 \times 67}{330}} = \sqrt{\frac{176 \times 67}{3300}} = \sqrt{\frac{44 \times 67}{825}} = \sqrt{\frac{2948}{825}}

t = \sqrt{\frac{268}{75}} \approx \sqrt{3.5733} \approx 1.890 s

The calculated time $t \approx 1.89 s$ is closest to $2 s$ among the given options.

💡 Teacher's Secret Hint

Pay close attention to unit consistency and significant figures. In multiple-choice questions, if your calculated value is very close to an option, that option is likely the intended answer.

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Physics Question 30 – JEE-MAIN 2026

Clue 1 of 3: Frame of Reference

🧠 Full Solution Path