Q.An astronaut accidentally gets separated out of his small spaceship accelerating in interstellar space at a constant rate of $100\ \text{m s}^{-2}$. What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.)

Concept and Intuition

When a body is thrown upward from Earth’s surface, its total mechanical energy (kinetic + gravitational potential) is conserved because only gravity (a conservative force) acts. The escape velocity $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$ is the minimum speed needed for the body to reach infinity with zero kinetic energy. Here, the launch speed is $\frac{2}{3} v_{\text{esc}}$, so the body will not escape; it will rise to a finite maximum height $h$ where its kinetic energy becomes zero. We equate the total energy at launch to the total energy at the highest point, using the correct gravitational potential energy formula (not $mgh$, because $h$ is comparable to Earth’s radius).

Step-by-step solution

1. Write the escape velocity formula Escape velocity from Earth’s surface:

where $G$ is the gravitational constant, $M$ is Earth’s mass, and $R = 6400$ km is Earth’s radius.

2. Given launch speed

3. Conservation of mechanical energy At the surface (point 1):

At maximum height $h$ (point 2): velocity = 0, distance from Earth’s center = $R + h$:

Set $E_1 = E_2$:

4. Cancel $m$ and substitute $v^2$

Simplify the first term:

So the equation becomes:

5. Combine the left side

Thus:

Cancel the minus signs and $GM$:

$\frac{5}{9R} = \frac{1}{R+h}$ …

The body has mass $m = W/g$. Since it moves rigidly with the helicopter, its acceleration components are

Only two forces act on the body: the tension $T$ (along the rope, at some angle $\theta$ to the vertical) and its weight $W$ acting straight down.

Step-by-step reasoning

1. Required net force.

2. Resolve the tension. Horizontally, only the tension acts:

Vertically, tension acts up and weight acts down, and their net must equal $F_{\text{net},y} = W$ (upward):

3. Solve for the tension. Squaring and adding:

4. Cross-check with vectors. The net force vector required is $\vec{F}_{\text{net}} = W\hat{i} + W\hat{j}$. The two actual forces are $\vec{T}$ and the weight $-W\hat{j}$, so …

Concept & Intuition

“Apparent weight” is not your true weight ($mg$), but the force you feel through the floor — the normal reaction $N$. When a lift accelerates, your body “wants” to keep moving at constant velocity (Newton’s first law), so the floor must push harder or less hard to change your motion.

Here the lift is moving up but slowing down (retardation = $2.8~\mathrm{m/s^2}$ upward). Slowing down while going up means acceleration is downward. So you feel lighter, because the floor doesn’t need to push as hard — part of gravity is “used” to decelerate you.

Step-by-step reasoning

1. Choose a sign convention. Let upward be positive. The lift’s velocity is upward, but it is retarding, so its acceleration is downward:

2. Draw the free-body diagram for the man.

   Two forces act:

   • Weight $W = mg$ downward (negative).
   • Normal reaction $N$ from the floor upward (positive).

   Newton’s second law:

3. Plug in the numbers. $m = 60~\mathrm{kg}$, $g = 9.8~\mathrm{m/s^2}$, $a = -2.8~\mathrm{m/s^2}$:

4. Solve for $N$. $N = 588 - 168 = 420~\mathrm{N}.$ …

Concept & Intuition

When two masses are connected by a string over a pulley, the heavier mass accelerates downward and the lighter one upward. The centre of mass (COM) of the system does not move with constant acceleration equal to either mass’s acceleration — instead, it’s the mass-weighted average of their accelerations. Since the accelerations are opposite in direction, the COM acceleration ends up being much smaller than either individual acceleration. The key is to first find each block’s acceleration using Newton’s laws, then combine them.

Step-by-step solution

1. Set up the masses and forces

   Let the masses be $m_1 = m$ and $m_2 = n$ (the ratio is $m:n$, but we can treat them as actual masses for calculation; the ratio will appear naturally). Assume $m > n$ so that $m$ moves downward and $n$ upward. Tension $T$ is the same on both sides (light, inextensible string, frictionless pulley).

2. Write Newton’s second law for each block

   For the heavier mass $m$ (downward positive):

For the lighter mass $n$ (upward positive):

Here $a$ is the magnitude of acceleration of each block (they move with the same magnitude but opposite directions).

3. Solve for the acceleration $a$ Add the two equations:

This is the familiar Atwood machine acceleration.

4. Find the acceleration of the centre of mass The centre of mass acceleration is the mass-weighted average of the individual accelerations. Take upward as positive.
   • Block $m$ accelerates downward: $a_m = -a$
   • Block $n$ accelerates upward: $a_n = +a$

Substitute $a = \frac{m - n}{m + n}\, g$:

Notice $n - m = -(m - n)$, so:

Concept & Intuition

The jumper first accelerates downward at $g = 10\ \mathrm{m/s^2}$ for a short distance (20 m). During this phase, speed increases from zero. Once the parachute opens, the problem states the motion becomes uniform — meaning the speed stays constant at whatever value was reached at the end of free fall. So we need to:

1. Find the speed after falling 20 m.
2. Find the time to fall that 20 m.
3. Then find the time to cover the remaining distance at that constant speed.
4. Add the two times.

A common pitfall is forgetting that the free‑fall distance is only 20 m, not the whole 2 km, and that the constant speed is the final speed from free fall, not an average.

Step‑by‑step solution

1. Free‑fall phase (first 20 m) Initial velocity $u = 0$, acceleration $a = g = 10\ \mathrm{m/s^2}$, distance $s_1 = 20\ \mathrm{m}$. Use $v^2 = u^2 + 2as$ to find the speed just before the parachute opens:

The time for this phase comes from $v = u + at$:

2. Uniform‑motion phase (remaining distance) Total height = 2 km = 2000 m. Already fallen 20 m, so remaining distance: $s_2 = 2000 - 20 = 1980\ \mathrm{m}.$ …

The problem involves a body on a rough surface, so friction is present and opposes motion. The applied force must overcome friction before producing acceleration. Since the surface is rough and the body is already moving, kinetic friction acts with a constant magnitude (as long as the normal force and surface properties don't change). This means the net force is the applied force minus the constant frictional force. To double the acceleration, we need to double the net force, not the applied force — that's the central insight.

Let's work through it step by step.

1. Find the frictional force using the first condition. Mass $m = 3\ \text{kg}$, applied force $F_1 = 20\ \text{N}$, acceleration $a_1 = 4\ \text{m/s}^2$. Newton's second law: $F_{\text{net}} = m a$. The net force is $F_1 - f$, where $f$ is the kinetic friction. So:

Solving: $f = 20 - 12 = 8\ \text{N}$.

2. Determine the net force needed for double the acceleration.

   Desired acceleration $a_2 = 2 \times 4 = 8\ \text{m/s}^2$.

   Required net force: $F_{\text{net,2}} = m a_2 = 3 \times 8 = 24\ \text{N}$.

3. Find the new applied force.

   The friction is still $8\ \text{N}$ (constant). So:

4. Calculate the percentage increase in the applied force. Increase = $F_2 - F_1 = 32 - 20 = 12\ \text{N}$. …

The key idea here is a classic result from Newton’s shell theorem: a uniform spherical shell exerts zero net gravitational force on any mass placed inside it. That means the unit mass at the centre of the shell feels nothing from the shell itself — the only contribution comes from the solid sphere.

But the solid sphere is not centred at the same point. Its centre is offset because the sphere and shell touch each other. So we need to find the distance between the centre of the shell (where our test mass sits) and the centre of the solid sphere.

1. Locate the centres. The spherical shell has radius $4R$, so its centre is at point O. The solid sphere has radius $R$, and its surface touches the inner surface of the shell. Since the shell’s inner radius is $4R$, the distance from O to the centre of the solid sphere (call it O') is the shell’s radius minus the sphere’s radius:

2. Force from the shell. By the shell theorem, a uniform spherical shell exerts zero gravitational field at any point inside it. Our unit mass is at the centre O, which is certainly inside the shell. Therefore,

3. Force from the solid sphere. For a point outside a uniform solid sphere, the sphere behaves as if all its mass were concentrated at its centre. The unit mass at O is at a distance $3R$ from the sphere’s centre O', so it is outside the sphere (since $3R > R$). Hence the force is simply …

Concept & Intuition

When the man jumps vertically down from the bridge, he has no horizontal velocity relative to the ground at the moment he leaves the bridge. The boat is moving horizontally at $10~\mathrm{m/s}$. As the man lands on the boat, the only horizontal forces are internal (between man and boat), so the total horizontal momentum of the man–boat system is conserved. The boat slows down because it must share its horizontal momentum with the man.

1. Define the system and initial conditions
   • Mass of man: $m = 60~\mathrm{kg}$
   • Mass of boat: $M = 540~\mathrm{kg}$
   • Initial speed of boat: $v_b = 10~\mathrm{m/s}$ (horizontal)
   • Initial speed of man (horizontal): $v_m = 0$ (he jumps straight down)
   • Total initial horizontal momentum:

2. After the man lands on the boat
   • The man and boat move together with a common horizontal speed $v_f$.
   • Total mass: $M + m = 540 + 60 = 600~\mathrm{kg}$
   • Final horizontal momentum:

3. Apply conservation of horizontal momentum

$v_f = \frac{5400}{600} = 9~\mathrm{m/s}$ …

Concept & Intuition

When a stream of water hits a block and rebounds with the same speed, the water’s momentum changes direction. That change in momentum per unit time is the force the water exerts on the block. Since the surface is smooth (no friction), this force is the net force on the block, giving it an acceleration. The key is to realize that the water’s velocity reverses, so the change in velocity is twice the original speed, not just once.

Step-by-step solution

1. Identify the relevant physics The force on the block comes from the water jet. For a fluid stream hitting a surface and rebounding, the force is given by the rate of change of momentum of the fluid:

Here, each water molecule’s velocity changes from $+v$ (toward the block) to $-v$ (away), so the change in velocity for each molecule is $\Delta v = v - (-v) = 2v$.

2. Find the mass flow rate of water The water comes from a pipe of cross-sectional area $A = 5\,\mathrm{cm}^2 = 5 \times 10^{-4}\,\mathrm{m}^2$ with speed $v = 5\,\mathrm{m/s}$. The volume of water hitting the block per second is $A v$. Density $\rho = 1\,\mathrm{g/cc} = 1000\,\mathrm{kg/m}^3$. So the mass flow rate (mass per second) is:

3. Calculate the force on the block The momentum change per second for the water is: $F = \frac{\Delta m}{\Delta t} \times (\text{change in velocity}) = (2.5\,\mathrm{kg/s}) \times (2 \times 5\,\mathrm{m/s}) = 2.5 \times 10 = 25\,\mathrm{N}$ …

Energy conservation from the surface to infinity (where potential energy is zero):

Using $\dfrac{GM}{R_E}=gR_E$ and the launch speed $v_0=\sqrt{4gR_E}$ (so $v_0^2=4gR_E$): …