JEE Physics - MCQ Practice Questions

Step 1: Differentiate the velocity function

The acceleration is the time derivative of velocity:

**Step 2: Evaluate at $t=2\text{s}$

Result:

Step 1: Recall the kinematic equation

At maximum height, the final velocity is $v=0$. We use the equation

where $u=20\text{m/s}$ is the initial velocity and $g=10{\text{m/s}}^2$.

**Step 2: Substitute and solve for $h$

Result

Step 1: Express momentum in terms of base quantities

Momentum is defined as the product of mass and velocity:

Step 2: Substitute SI base units

The SI unit of mass is $\text{kg}$ and the SI unit of velocity is $\text{m/s}$, so:

Step 3: Note on equivalent units

$\text{N⋅s}$ (newton-second) is dimensionally equivalent to $\text{kg⋅m/s}$, since $1\text{N}=1{\text{kg⋅m/s}}^2$, giving:

However, $\text{N⋅s}$ is a derived combination and is not the fundamental SI expression for momentum.

Result:

The SI unit of momentum is:

Step 1: Write Newton's Second Law

From Newton's second law, $F=ma$, so the acceleration is

Step 2: Set up the ratio of accelerations

Since the force $F$ is the same for both bodies, the acceleration is inversely proportional to the mass:

Step 3: Substitute the mass ratio

Given $m_A:m_B=1:2$, we have

Result

The ratio of their accelerations is

Step 1: Find the acceleration

Using the formula $a=\frac{v-u}{t}$:

Step 2: Find the distance using kinematics

Using $s=ut+\frac{1}{2}a{t}^2$:

Result:

Step 1: Identify the formula for maximum static friction.

Step 2: Substitute the normal force $N=mg$ for a block on a horizontal surface.

Step 3: Substitute the known values $\mu =0.2$, $m=5\text{kg}$, and $g=10{\text{m/s}}^2$.

Step 1: Recall the range formula

The horizontal range of a projectile is given by

Step 2: Substitute the known values

Here $u=40\text{m/s}$, $\theta =45°$, and $g=10{\text{m/s}}^2$.

Step 3: Final Answer

Step 1: Recall the elastic collision formula

For a perfectly elastic collision, the post-collision velocity of mass $m_1$ is given by

Step 2: Identify the known values

Step 3: Substitute and simplify

Result

The velocity of the $2\text{kg}$ mass after the collision is

Step 1: Recall Newton's Second Law

Newton's second law states:

Step 2: Solve for mass

Rearranging for $m$:

Step 3: Substitute the known values

With $F=10\text{N}$ and $a=2{\text{m/s}}^2$:

Newton's First Law

A body moving with uniform velocity has a constant velocity, which means its acceleration is zero:

By Newton's Second Law:

Substituting $a=0$:

Therefore, the net force on the body is:

Step 1: Write the tension formula

The tension in the rope is related to the weight and horizontal force by

Step 2: Substitute known values

The weight of the mass is $mg=40\times 10=400\text{N}$, and the maximum tension is $T=500\text{N}$. Substituting:

**Step 3: Solve for $F^2$

Squaring both sides:

**Step 4: Solve for $F$

Step 1: Define the velocity direction

For a particle in circular motion, the velocity $\mathbf{v}$ is always tangent to the circle, pointing along the circumference at any given instant.

Step 2: Define centripetal acceleration

The centripetal acceleration ${\mathbf{a}}_c$ is defined as:

where $v$ is the speed, $r$ is the radius of the circle, and $\hat{r}$ is the unit vector pointing radially outward from the center. The negative sign indicates the acceleration points radially inward.

**Step 3: Relationship between ${\mathbf{a}}_c$ and $\mathbf{v}$

Since $\mathbf{v}$ is tangential and ${\mathbf{a}}_c$ is radial, they satisfy:

confirming that ${\mathbf{a}}_c$ is perpendicular to $\mathbf{v}$ at every point on the path.

Conclusion

The centripetal acceleration is directed:

Step 1: Find the system acceleration

The total mass of the system is $m_1+m_2=3+2=5\text{kg}$. Applying Newton's second law to the entire system:

Step 2: Find the tension in the string

The tension $T$ accelerates block $m_2$ alone:

Result:

Step 1: Find time to maximum height

Since the ball returns to the same height in $4\text{s}$, by symmetry the time to reach maximum height is

Step 2: Find initial velocity

At maximum height the velocity is zero, so using $v=u-gt$ with $v=0$:

Step 1: Establish the original range formula

For a projectile launched horizontally from height $h$ with initial velocity $u$, the time to fall is found from:

The horizontal range is therefore:

Step 2: Apply the new conditions

The new height is $2h$ and the new velocity is $2u$. The new range $x^{\prime }$ is:

**Step 3: Express the new range in terms of $x$

Rewrite $4u\sqrt{\frac{h}{g}}$ to match the original expression $u\sqrt{\frac{2h}{g}}$:

Result

N - mg = ma. N = m(g+a) = 50(10+2) = 600 N

Centripetal force is perpendicular to velocity (displacement), so W = F·s·cosθ = F·s·cos90° = 0

Without friction: v² = 2gh. With friction and 2h: friction opposes motion, so final velocity < √(4gh) = √2v

v = dx/dt = 8t - 2. At t = 1s, v = 8(1) - 2 = 6 m/s

For equal masses in elastic collision with one at rest: v₁' = 0, v₂' = u₁ (velocities exchange)