Q.A shell of mass $0.020\ \text{kg}$ is fired by a gun of mass $100\ \text{kg}$. If the muzzle speed of the shell is $80\ \text{m s}^{-1}$, what is the recoil speed of the gun?

Concept & Intuition

Linear momentum is defined as $p = mv$. If $p$ is constant, then $v = p/m$. This is an inverse relationship: as mass increases, velocity decreases, and vice versa. The graph of $v$ vs. $m$ is therefore a rectangular hyperbola — a curve that falls steeply at small masses and flattens out, approaching zero as mass grows large, but never actually reaching zero. Among the given options, only one shows this characteristic shape.

Step-by-step reasoning

1. Write the relation Constant momentum means $p = \text{constant}$. Hence

This is of the form $v = k/m$ with $k = p > 0$.

2. Identify the graph type

   The equation $v = k/m$ is a rectangular hyperbola. Its key features:

   • As $m \to 0^+$, $v \to \infty$ (vertical asymptote at $m=0$).
   • As $m \to \infty$, $v \to 0$ (horizontal asymptote at $v=0$).
   • The curve is decreasing and convex (curving downward) for all $m > 0$.

3. Match with the options

   • (A) shows a curve that starts high near the velocity axis, drops steeply, then flattens and approaches the mass axis — exactly a hyperbola.
   • (B) is a straight line with negative slope — that would mean $v$ decreases linearly with $m$, which is not $v \propto 1/m$.
   • (C) is a concave-down arc that meets the mass axis at a finite point — a hyperbola never touches the mass axis (it only approaches it).
   • (D) is a straight line through the origin with positive slope — that would mean $v \propto m$, which is the opposite of the required relation. …

Concept & Intuition

When an object at rest explodes, no external horizontal forces act (we ignore air resistance and gravity’s effect on horizontal motion). Therefore, the total momentum before and after the explosion must be equal. Before the explosion, momentum is zero. After, the two fragments fly apart in opposite directions (or at least with opposite velocity components) so that their vector sum remains zero. Once we know the velocity of one piece, we can find the other’s velocity from momentum conservation, and then kinetic energy follows directly.

1. State the given data

   • Total mass: $M = 20\ \text{kg}$
   • Mass of first fragment: $m_1 = 12\ \text{kg}$
   • Mass of second fragment: $m_2 = 8\ \text{kg}$
   • Velocity of the 8 kg fragment: $v_2 = 6\ \text{m/s}$ (direction is not needed for magnitude, but we’ll treat it as positive; the other fragment will move opposite).

2. Apply conservation of momentum

   Initial momentum = $0$.

   Final momentum: $m_1 v_1 + m_2 v_2 = 0$.

   So

The negative sign means the 12 kg fragment moves in the opposite direction to the 8 kg fragment. Its speed is $4\ \text{m/s}$.

3. Compute the kinetic energy of the 12 kg mass …

The key idea is that momentum is defined as $p = mv$. If momentum is held constant, then $v = \frac{p}{m}$. This is an inverse relationship: as mass increases, velocity decreases, and vice versa. The graph of an inverse proportion is a rectangular hyperbola — not a straight line.

Let’s work through the options step by step.

1. Identify the relationship

   Constant momentum means $p = \text{constant}$.

   So $v = \frac{p}{m}$. This is of the form $v = \frac{k}{m}$ where $k = p$.

   This is an inverse variation: when $m$ is small, $v$ is large; when $m$ is large, $v$ is small. The curve never touches either axis (as $m \to 0$, $v \to \infty$; as $m \to \infty$, $v \to 0$).

2. Examine each figure

   • Fig 1: A smooth curve that starts high on the $v$-axis at small $m$, falls steeply, then flattens and approaches the $m$-axis asymptotically. This is exactly the shape of $v = k/m$ — a rectangular hyperbola.
   • Fig 2: A straight line with constant negative slope. That would mean $v$ decreases linearly with $m$, i.e., $v = a - bm$. This is not an inverse relationship.
   • Fig 3: A straight line through the origin with positive slope. That means $v \propto m$, i.e., direct proportion. This would imply momentum increases with mass, not constant.
   • Fig 4: A smooth curve that rises slowly at first, then more steeply — concave up. This suggests $v$ increases faster than $m$ (e.g., $v \propto m^2$), which is the opposite of what we need.

3. Match the correct figure …

For a system of particles, $\vec F_{ext,\text{net}}=\dfrac{d\vec P}{dt}$. If $\vec F_{ext,\text{net}}=0$, then $\vec P=$ constant, so linear momentum is conserved. …

In a one-dimensional elastic collision, the fraction of the incident particle's kinetic energy transferred to a stationary target is

With $m_1 = 1$ (neutron), $m_2 = 206$ (lead nucleus): …

The key idea is to break the motion into three clean stages:

1. Swing down – the left bob converts gravitational potential energy into kinetic energy.
2. Collision – the two bobs stick together; momentum is conserved, but kinetic energy is lost.
3. Swing up – the combined bobs convert their kinetic energy back into gravitational potential energy.

Because the bobs stick together, this is a perfectly inelastic collision. That means we cannot simply average the speeds; we must use momentum conservation to find the speed just after impact, then use energy conservation to find the height.

Step-by-step reasoning

1. Speed of the left bob just before collision The left bob is released from rest at height $h$ above the lowest point. By conservation of mechanical energy (no friction, strings are ideal):

This is the speed of the left bob just before it hits the stationary right bob.

2. Momentum conservation during the collision Both bobs have the same mass $m$. The right bob is initially at rest. After the collision they stick together, so the combined mass is $2m$ and they move with a common speed $V$. Momentum before = momentum after:

Substituting $v = \sqrt{2gh}$:

3. Height reached after collision After the collision, the two bobs (mass $2m$) swing upward together. Their kinetic energy just after collision is converted entirely into gravitational potential energy at the maximum height $H$:

Simplify:

[!TLDR]

Conserve momentum during the embedding, then equate the post-collision kinetic energy to $(m+M)gh$; the result has the $\sqrt{2gh}\,(1+\text{mass ratio})$ form of option (B).

Concept

This is the ballistic-pendulum problem (CBSE Class-11 Work, Energy and Collisions). The collision is perfectly inelastic — momentum is conserved but kinetic energy is not — after which mechanical energy of the combined block-plus-bullet is conserved as it rises.

Solution

Let the bullet (mass $m$, speed $v$) embed in the block (mass $M$), giving common speed $V$.

Momentum conservation during the collision:

Energy conservation as the combined mass rises to height $h$:

Combining:

$\frac{m v}{m+M}=\sqrt{2gh} \;\Rightarrow\; v=\frac{m+M}{m}\sqrt{2gh}=\sqrt{2gh}\left(1+\frac{M}{m}\right).$ …

Step 1 — Test (A): momentum of the ball alone.

During the bounce the floor pushes up on the ball with a large normal force $\vec{N}$ for a time $\Delta t$. That is an external force on the ball, delivering an impulse

Indeed the ball's momentum literally reverses direction (down → up). So (A) is false.

Step 2 — Test (C): momentum of the ball + Earth system.

Now the ball–floor contact force and the ball–Earth gravitational force are internal (Newton's third-law pairs inside the system: the ball pushes down on the Earth exactly as hard as the Earth pushes up on the ball). With no external force,

The Earth recoils imperceptibly (its mass is $\sim 10^{25}$ times larger), but its momentum change exactly cancels the ball's. So (C) is true. …

Step 1 — List every real force on the particle.

1. Weight $mg$, vertically down.
2. Normal reaction $N$ from the table, vertically up.
3. Tension $T$ in the string, horizontal, directed from the particle towards the peg — i.e. towards the centre of the circle.

The table is smooth, so there is no friction, and the motion is horizontal, so

The vertical forces cancel exactly and contribute nothing to the horizontal (radial) direction.

Step 2 — Apply Newton's second law along the radius.

For circular motion the net radial (centripetal) force must supply the required centripetal acceleration $a_c = v^2/l$:

But the only force with a component towards the centre is $T$. Hence

Step 3 — Why the other options are traps. …