Q.A cricket ball of mass 150 g has an initial velocity $\mathbf{u} = (3\hat{i} + 4\hat{j})$ m s$^{-1}$ and a final velocity $\mathbf{v} = -(3\hat{i} + 4\hat{j})$ m s$^{-1}$ after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s$^{1}$)

In projectile motion the horizontal velocity component stays constant, while the vertical component decreases with height. At the top the vertical component is zero; at any lower height it is found from a kinematic equation.

1. Launch components ($\theta = 45^\circ$):

2. Speed at maximum height ($v_y = 0$):

3. Maximum height and half of it

4. Vertical velocity at half height ($v_y^2 = u_y^2 - 2gh$):

5. Speed at half height

6. Required ratio

i.e. $v_{\text{max}} : v_{\text{half}} = \sqrt2 : \sqrt3$.

The concept here is simple but often missed: work is a scalar, computed as the dot product of the force vector and the displacement vector. That dot product picks out the component of force in the direction of motion — any force perpendicular to displacement does zero work.

In this problem, the displacement is $(2\sqrt{3}\,\hat{i})$ m — it has no $\hat{j}$ component at all. So the $\hat{j}$-component of the force, however large, does no work. Only the $\hat{i}$-component matters.

1. Write the force vector:

   $\vec{F} = (6\sqrt{3}\,\hat{i} - 4\sqrt{2}\,\hat{j})$ N

2. Write the displacement vector:

   $\vec{s} = (2\sqrt{3}\,\hat{i})$ m

3. Work done is $W = \vec{F} \cdot \vec{s}$.

   Dot product: multiply the $\hat{i}$-components together, multiply the $\hat{j}$-components together, and add.

   $W = (6\sqrt{3})(2\sqrt{3}) + (-4\sqrt{2})(0)$

4. Simplify:

   $(6\sqrt{3})(2\sqrt{3}) = 6 \times 2 \times (\sqrt{3} \times \sqrt{3}) = 12 \times 3 = 36$

   The $\hat{j}$ term is zero.

So $W = 36$ J.

Convert data: m = 250 g = 0.25 kg, v = 72 km/h = 72 x 5/18 = 20 m/s. The ball is turned through theta = 120 deg without change in speed.

Impulse equals the change in momentum. For two vectors of equal magnitude p = mv making an angle theta between them, the magnitude of their difference is

ANSWER: B

The key idea is that on a banked curve, there are two special speeds:

• Optimum speed $v_{\text{opt}}$: the speed at which no friction is needed — the horizontal component of the normal force alone provides the centripetal force.
• Maximum permissible speed $v_{\max}$: the fastest speed before the car slips outward, where static friction acts down the bank (opposing the tendency to slide up) and is at its maximum value $\mu_s N$.

The problem tells us $v_{\max} = 2 v_{\text{opt}}$. We can write the force equations for both cases and solve for $\mu_s$.

1. Optimum speed (no friction) For a banked curve at angle $\theta = 45^\circ$, the normal force $N$ has a vertical component $N \cos\theta$ balancing weight, and a horizontal component $N \sin\theta$ providing centripetal force:

Dividing the second by the first gives:

With $\theta = 45^\circ$, $\tan 45^\circ = 1$, so:

2. Maximum speed (friction at its limit) At maximum speed, the car is about to slide up the bank, so friction acts down the bank (opposing the upward slide). The friction force is $f = \mu_s N$, directed down the incline. Resolve forces horizontally (toward the center) and vertically:
   • Horizontal (centripetal):

• Vertical (no acceleration):

Substitute $f = \mu_s N$ and $\theta = 45^\circ$ ($\sin 45^\circ = \cos 45^\circ = 1/\sqrt{2}$):

3. Eliminate $N$ and use the given ratio Divide equation (1) by equation (2):

From step 1, $R g = v_{\text{opt}}^2$. The problem states $v_{\max} = 2 v_{\text{opt}}$, so $v_{\max}^2 = 4 v_{\text{opt}}^2 = 4 R g$. Thus:

4. Solve for $\mu_s$

Concept & Intuition

This problem is about projectile motion, but we don’t need to break it into horizontal and vertical components or worry about the angle. Why? Because the only force doing work is gravity (a conservative force), and the ball’s speed at any point depends only on the vertical distance it has fallen or risen — not on the direction of motion. The angle only tells us the initial vertical velocity, but the final speed condition gives us a direct energy relation. So we use conservation of mechanical energy: the sum of kinetic and gravitational potential energy is constant (ignoring air resistance).

Step-by-step solution

1. Identify the given data

   Initial speed: $u = 40\ \mathrm{m/s}$

   Final speed: $v = 1.25 \times u = 1.25 \times 40 = 50\ \mathrm{m/s}$

   Acceleration due to gravity: $g = 10\ \mathrm{m/s^2}$

   Initial height above ground: $h$ (unknown)

   Angle of projection: $30^\circ$ (irrelevant for energy)

2. Set up the energy conservation equation

   Take the ground as zero potential energy.

   Initial energy (at height $h$, moving with speed $u$):

Final energy (just before striking the ground, height = 0, speed = $v$):

Since no non-conservative forces act, $E_i = E_f$.

3. Cancel mass and solve for $h$

Divide through by $m$:

Rearrange:

So:

4. Substitute the numbers

The core idea here is understanding what "breaking off the plane" means in terms of forces. A body resting on a horizontal plane experiences a normal force from the plane, supporting its weight. If an upward force is applied, it reduces the normal force. When this upward force becomes equal to the body's weight, the normal force becomes zero, and the body loses contact with the plane.

The applied force $F = \alpha t$ acts at an angle, so it has both a vertical component that affects the normal force and a horizontal component that causes the body to accelerate horizontally. We need to find the specific time when the normal force vanishes, and then calculate the horizontal velocity accumulated up to that moment.

1. Identify and Resolve Forces:

   The body has a mass $m = 1$ kg. It is on a smooth horizontal plane, meaning there is no friction. The forces acting on the body are:

   • Its weight, $W = mg$, acting vertically downwards. We will use $g = 10 \text{ m/s}^2$ as is standard in such problems unless specified otherwise. So, $W = (1 \text{ kg})(10 \text{ m/s}^2) = 10 \text{ N}$.
   • The normal force, $N$, exerted by the plane, acting vertically upwards.
   • The applied force, $F = \alpha t$, acting at an angle of $45^\circ$ with the horizontal.

   We resolve the applied force into its horizontal ($F_x$) and vertical ($F_y$) components:

   • Horizontal component: $F_x = F \cos 45^\circ = \alpha t \cdot \frac{1}{\sqrt{2}}$.
   • Vertical component: $F_y = F \sin 45^\circ = \alpha t \cdot \frac{1}{\sqrt{2}}$.

2. Determine the Condition for Breaking Off:

   The body breaks off the plane when it loses contact with the surface. This happens when the normal force $N$ exerted by the plane on the body becomes zero.

   Consider the forces in the vertical direction. Taking the upward direction as positive:

At the moment the body breaks off, $N = 0$. Therefore, the condition for breaking off is:

3. Calculate the Time ($t_0$) When the Body Breaks Off: Substitute the expression for $F_y$ and the value of $mg$ into the condition $F_y = mg$:

Solving for $t_0$:

This is the time at which the body lifts off the plane.

4. Calculate the Horizontal Acceleration:

While the body is on the plane (for $t \le t_0$), the horizontal component of the applied force causes horizontal acceleration. Since the plane is smooth, there is no friction. According to Newton's second law ($F = ma$):

Substitute $F_x = \alpha t \cdot \frac{1}{\sqrt{2}}$ and $m = 1$ kg:

5. Calculate the Velocity at Time $t_0$: The body starts from rest, so its initial velocity $v(0) = 0$. The velocity at time $t_0$ is found by integrating the acceleration from $t=0$ to $t=t_0$:

6. Substitute $t_0$ and Simplify: Now, substitute the value of $t_0 = \frac{10\sqrt{2}}{\alpha}$ into the expression for $v(t_0)$:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

Concept and Intuition

Momentum is a fundamental concept in physics that describes the "quantity of motion" an object possesses. It is defined as the product of an object's mass and its velocity.

Here, $p$ is momentum, $m$ is mass, and $v$ is velocity.

Crucially, momentum is a vector quantity. This means it has both magnitude and direction. When we talk about a change in momentum, we are talking about a vector difference:

The subtraction of vectors means we must account for their directions. If an object reverses its direction, its velocity vector changes sign, which significantly impacts the change in momentum.

Step-by-step Derivation

1. Define the initial momentum:

   Let's choose the direction of the body's initial motion (towards the wall) as the positive direction.

   The initial speed is $v$.

   Therefore, the initial velocity is $v_{\text{initial}} = +v$.

   The initial momentum is $p_{\text{initial}} = M v_{\text{initial}} = M(+v) = Mv$.

2. Define the final momentum:

   The problem states that the body "retraces its path with the same speed". This means its speed is still $v$, but its direction of motion is now opposite to the initial direction (away from the wall).

   Therefore, the final velocity is $v_{\text{final}} = -v$.

   The final momentum is $p_{\text{final}} = M v_{\text{final}} = M(-v) = -Mv$.

3. Calculate the change in momentum:

   Using the formula for change in momentum:

   $\Delta p = p_{\text{final}} - p_{\text{initial}}$

   Substitute the values we found:

   $\Delta p = (-Mv) - (Mv)$

   $\Delta p = -2Mv$

   Watch out

   A common mistake is to treat momentum as a scalar quantity and simply subtract the magnitudes, leading to $Mv - Mv = 0$. This is incorrect because momentum is a vector, and the change in momentum must account for the change in direction.

4. Interpret the result:

   The result $\Delta p = -2Mv$ tells us two things:

   • The magnitude of the change in momentum is $2Mv$.
   • The negative sign indicates that the direction of the change in momentum is opposite to our chosen positive direction (i.e., opposite to the initial direction of motion, or in the direction away from the wall).

   In multiple-choice questions, when both $2Mv$ and $-2Mv$ are not simultaneously available as options, or when the question asks for "the change in momentum" without specifying a direction, the magnitude of the change is often implied. If we had chosen the final direction (away from the wall) as positive, the initial velocity would be $-v$ and the final velocity $+v$, leading to $\Delta p = Mv - (-Mv) = 2Mv$. Both $2Mv$ and $-2Mv$ represent the same physical change, just described relative to different positive reference directions. Given the options, $2Mv$ is the intended answer, representing the magnitude of this change.

The core concept here is Newton's Second Law of Motion, which relates force to the rate of change of momentum. When a force acts on an object, it changes the object's momentum. If this force is constant, the change in momentum is simply the product of the force and the time for which it acts. If the force varies, we can talk about an average force over a given time interval.

The intuition is straightforward: to stop a moving object, you need to apply a force that opposes its motion. The heavier the object, or the faster it's moving, the more momentum it has, and thus a larger force (or a longer time) is required to bring it to rest. The problem provides all the necessary information to calculate this average force directly.

1. Identify the given quantities:

   • Mass of the ball, $m = 0.2\,\text{kg}$
   • Initial speed of the ball, $u = 20\,\text{m/s}$
   • Final speed of the ball, $v = 0\,\text{m/s}$ (since it is brought to rest)
   • Time taken to stop the ball, $t = 0.1\,\text{s}$

2. Recall the relationship between average force and momentum:

   Newton's second law states that the net force acting on an object is equal to the rate of change of its momentum. For an average force, this can be written as:

   $F_{\text{avg}} = \frac{\Delta p}{\Delta t}$

   where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval.

   Momentum $p$ is defined as the product of mass and velocity ($p = mv$). So, the change in momentum $\Delta p$ is $m(v - u)$.

3. Substitute the values into the formula:

   The change in momentum is:

   $\Delta p = m(v - u)$

   $\Delta p = 0.2\,\text{kg} \times (0\,\text{m/s} - 20\,\text{m/s})$

   $\Delta p = 0.2\,\text{kg} \times (-20\,\text{m/s})$

   $\Delta p = -4\,\text{kg}\cdot\text{m/s}$

   Now, calculate the average force:

   $F_{\text{avg}} = \frac{\Delta p}{t}$

   $F_{\text{avg}} = \frac{-4\,\text{kg}\cdot\text{m/s}}{0.1\,\text{s}}$

   $F_{\text{avg}} = -40\,\text{N}$

   The negative sign indicates that the force is in the direction opposite to the initial motion of the ball, which is expected for a force that brings the ball to rest. The question asks for the magnitude of the average force.

4. Determine the magnitude of the force:

   The magnitude of the average force is $|-40\,\text{N}| = 40\,\text{N}$.