Q.A bob of mass $m$ is suspended by a light string of length $L$. It is imparted a horizontal velocity $v_o$ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. Obtain an expression for

Concept and Intuition

When a ball is dropped from a height, it gains kinetic energy as it falls. On hitting the ground, it rebounds with a speed reduced by the coefficient of restitution $e$. This means the height of each successive bounce is $e^2$ times the previous height. The question asks: Between the second and third collisions, when does the linear momentum become zero? That happens at the highest point of the second bounce — because at the peak, the ball momentarily stops before falling again. At that point, its kinetic energy is zero, so all its mechanical energy is gravitational potential energy. We need to find what percentage of the original potential energy (from point A) is lost by the time the ball reaches that peak.

Step-by-step solution

1. Initial potential energy The ball is dropped from height $h_0 = 10 \, \text{m}$, mass $m = 0.5 \, \text{kg}$. Initial potential energy:

(We can keep $g$ symbolic; it will cancel.)

2. Height after first bounce Coefficient of restitution $e = 0.5$. After the first impact, the rebound speed is $e$ times the impact speed. Since height is proportional to the square of speed, the height after the first bounce is:

3. Height after second bounce The same factor applies again: after the second impact, the ball rebounds to a height:

4. Where does momentum become zero between second and third collisions? The ball is in the air between the second and third collisions. Its momentum becomes zero at the top of the second bounce, i.e., at height $h_2$. At that point, the ball has only potential energy: $$$$

Given: length $L = 1\ \text{m}$, breadth $b = 2.5\ \text{cm} = 2.5\times10^{-2}\ \text{m}$, thickness $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$, central load $W = 25\ \text{N}$, and $Y = 2\times10^{11}\ \text{N m}^{-2}$.

Depression of a centrally-loaded beam supported at both ends:

Substitute the values:

$d^3 = (5\times10^{-3})^3 = 1.25\times10^{-7}\ \text{m}^3$ …

The key here is that the retarding force is not constant — it depends on position $x$. When a force varies with position, you cannot use $F = ma$ with constant acceleration. Instead, the work–energy theorem is the natural tool: the net work done on the block equals its change in kinetic energy. That theorem holds for any force, constant or variable, as long as you can compute the work integral.

The force is given as $F = -25x$, where the negative sign means it opposes the motion. The block moves from $x = 1.5$ m to $x = 2.5$ m, so the work done by this force is the integral of $F$ over that displacement.

1. Write the work–energy relation. The work done by the retarding force is

This work equals the change in kinetic energy:

Here $v_i = 6$ m/s and $v_f = 4$ m/s.

2. Evaluate the integral.

Compute the bracket:

So $W = -25 \times 2 = -50$ J.

The negative sign makes sense — the force opposes motion, so it removes energy from the block.

3. Set up the kinetic energy change. …

The key idea here is conservation of mechanical energy in a spring-block system on a frictionless surface. Since there is no friction, the total energy (kinetic + elastic potential) remains constant. At the maximum displacement (amplitude), all energy is stored as spring potential energy. At any other position, the kinetic energy is simply the total energy minus the potential energy at that position.

Let’s work through it step by step.

1. Identify the given data and convert to SI units.

   Spring constant: $k = 100 \, \text{N/m}$

   Amplitude (maximum stretch from equilibrium): $A = 8 \, \text{cm} = 0.08 \, \text{m}$

   Position of interest: $x = 3 \, \text{cm} = 0.03 \, \text{m}$ from the mean position.

2. Find the total mechanical energy of the system.

   At the extreme position ($x = A$), the block is momentarily at rest, so kinetic energy is zero. All energy is elastic potential energy:

3. Find the elastic potential energy at $x = 0.03 \, \text{m}$.

The key concept is energy conservation in a frictionless system. The block collides with the spring, compresses it, and then the spring pushes the block back. Since there's no friction, mechanical energy is conserved: the block's kinetic energy transforms entirely into spring potential energy at maximum compression, and then back into kinetic energy as the spring expands. The maximum momentum after the collision occurs when the block leaves the spring with its original speed (but opposite direction), so we just need to find the initial momentum from the given compression.

1. At maximum compression, the block is momentarily at rest, so all its initial kinetic energy has been converted into elastic potential energy of the spring. Let the initial speed of the block be $v_0$. Then:

Cancel the $\frac{1}{2}$:

2. Solve for the initial speed:

3. The initial momentum is: $p_0 = M v_0 = M \cdot L \sqrt{\frac{K}{M}} = L \sqrt{MK}$ …

Both particles have equal mass $m$. Take the initial speed of A as $10~\text{m/s}$ and B at rest.

Momentum conservation:

Kinetic energy (decreased by $1\%$, so final $= 0.99$ of initial):

From these, $2 v_A v_B = (v_A+v_B)^2 - (v_A^2+v_B^2) = 100 - 99 = 1$, so $v_A v_B = 0.5$.

Thus $v_A,\,v_B$ are roots of $t^2 - 10t + 0.5 = 0$:

$t = \frac{10 \pm \sqrt{100-2}}{2} = \frac{10 \pm \sqrt{98}}{2} \approx 9.95 \text{ or } 0.05$ …

Concept and Intuition

When a gun fires a shell, the total momentum of the gun + shell system is conserved if no external horizontal forces act (like friction). Before firing, both are at rest, so total momentum = 0. After firing, the shell moves forward and the gun recoils backward. Their momenta must be equal in magnitude and opposite in direction:

The key trap: units. The shell’s speed is given in km/min, but the answer choices are in cm/s. We must convert everything to a consistent system (e.g., kg, m, s or g, cm, s).

Step-by-step solution

1. Write down given data

   • Mass of shell: $m_s = 20\ \text{g}$
   • Mass of gun: $m_g = 100\ \text{kg} = 100\,000\ \text{g}$ (since 1 kg = 1000 g)
   • Muzzle speed of shell: $v_s = 6\ \text{km/min}$

2. Convert shell speed to cm/s

   • $1\ \text{km} = 1000\ \text{m} = 100\,000\ \text{cm}$
   • $1\ \text{min} = 60\ \text{s}$
   • So $v_s = 6 \times \frac{100\,000\ \text{cm}}{60\ \text{s}} = 6 \times \frac{100\,000}{60} = 6 \times \frac{10\,000}{6} = 10\,000\ \text{cm/s}$ (Check: $6 \times 100\,000 / 60 = 600\,000 / 60 = 10\,000$)

3. Apply conservation of momentum

   Initial momentum = 0.

   Final momentum: $m_s v_s + m_g v_g = 0$ (taking forward as positive, so $v_g$ will be negative).

   Magnitude relation:

Solve for $|v_g|$:

4. Simplify …

The key idea is energy conservation: when the car collides with the spring and comes to rest, all its kinetic energy has been transferred into the spring’s stored elastic potential energy. No energy is lost (ideal spring, no friction), so we can set $\frac{1}{2} m v^2 = \frac{1}{2} k x^2$ and solve for $x$, the maximum compression.

1. Write the energy conservation equation The car’s kinetic energy just before impact is

The spring’s potential energy at maximum compression $x$ is

Because the car comes to rest, all kinetic energy becomes spring potential energy:

2. Cancel the common factor $\frac{1}{2}$ This simplifies to

3. Solve for $x$

4. Plug in the given values $m = 1000$ kg, $v = 10$ m/s, $k = 10^4$ N/m …

Concept & Intuition

When the car hits the spring, it slows down as the spring compresses. If we ignore friction and other losses, the car’s kinetic energy just before impact is entirely stored as elastic potential energy in the spring at the moment of maximum compression (when the car momentarily stops). This is a classic conservation of mechanical energy problem:

where $x$ is the maximum compression. The mass $m$, speed $v$, and spring constant $k$ are given, so we can solve directly for $x$.

Step-by-step solution

1. Write the energy conservation equation The car’s kinetic energy before collision:

The spring’s potential energy at maximum compression:

Setting them equal (no energy loss):

2. Cancel the common factor $\frac{1}{2}$

3. Solve for $x$

4. Plug in the numbers …

Launch speed from the spring

Convert the stored spring energy to kinetic energy ($m=0.2~\text{kg}$, $k=50~\text{N/m}$, $x=0.10~\text{m}$):

Projectile off the edge of the slab

The block leaves the slab horizontally at height $h=2~\text{m}$. Time to fall: …