How Forces Affect MotionClass 9 Science NCERT Solutions

According to Newton's first law of motion, an object moves with a constant velocity if the net force acting on it is zero. In this case, the table is moving at a constant velocity, which means its acceleration is zero.

Two horizontal forces are acting on the table:

For the net force to be zero, these two forces must be equal in magnitude and opposite in direction. $F_{net} = F - f = 0$ $f = F$

Therefore, the frictional force exerted by the floor on the table is equal in magnitude to the applied force F.

Let's analyze the forces on each block:

Block P: Two forces are acting on block P in opposite directions: 5 N and 4 N. The net force on block P is the difference between the two forces: $F_{net, P} = 5 \text{ N} - 4 \text{ N} = 1 \text{ N}$ Since the net force is not zero ($1 \text{ N} \neq 0$), block P experiences a net force and will accelerate.

Block Q: Block Q is moving with a constant velocity. According to Newton's first law of motion, if an object moves with a constant velocity, its acceleration is zero, and the net force acting on it must be zero. $F_{net, Q} = 0$ Therefore, Q does not experience a net force.

Combining these two findings, P experiences a net force, and Q does not experience a net force.

Correct statement: (i) P experiences a net force and Q does not experience a net force.

Given: Total oarsmen = 100 Number of oarsmen rowing forward, $N_f = 95$ Number of oarsmen rowing backward, $N_b = 5$ Force per oarsman, $F_{oar} = 200 \text{ N}$

To Find: Net force on the snake boat, $F_{net}$.

Calculation: When an oarsman rows backward, they push the water backward, and by Newton's third law, the water pushes the boat forward. Force propelling the boat forward, $F_{forward} = N_f \times F_{oar} = 95 \times 200 \text{ N} = 19000 \text{ N}$.

The 5 oarsmen rowing in the opposite direction push the water forward, which in turn pushes the boat backward. Force pushing the boat backward, $F_{backward} = N_b \times F_{oar} = 5 \times 200 \text{ N} = 1000 \text{ N}$.

The net force on the boat is the difference between the forward and backward forces. $F_{net} = F_{forward} - F_{backward}$ $F_{net} = 19000 \text{ N} - 1000 \text{ N} = 18000 \text{ N}$

Final Answer: The net force on the snake boat is 18000 N in the forward direction.

Newton's second law of motion states that when a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. Mathematically, $a = \frac{F_{net}}{m}$ From this relationship, we can see that:

Therefore, the correct statement is (iv).

Correct statement: (iv) in the direction of force, with acceleration proportional to the force acting on the object.

A net force acting on an object causes it to accelerate, which means its velocity changes over time. In a position-time graph, the slope of the line represents the velocity of the object.

Therefore, a net force acts on Object D.

Correct option: (iv) Object D

Yes, the boat will move.

Direction: The boat will move backward, away from the shore.

Reason: This phenomenon is explained by Newton's third law of motion. To jump forward, the sailor must push off from the boat. The sailor applies a force on the boat in the backward direction (this is the 'action' force). According to Newton's third law, the boat simultaneously exerts an equal and opposite force on the sailor in the forward direction (this is the 'reaction' force). This reaction force propels the sailor to the shore. The action force exerted on the boat causes it to accelerate and move backward.

The reason for placing a landing mat or sand bed is to reduce the force of impact on the athlete and prevent injury. This can be explained using Newton's second law of motion.

When the athlete lands, their velocity must be reduced to zero. This change in velocity means a change in momentum. The force of impact is related to the rate of change of momentum. According to Newton's second law, Force = (Change in momentum) / (Time of impact).

A hard surface like concrete would bring the athlete to a stop in a very short time. A short time of impact results in a very large force, which can cause serious injury.

The soft landing mat or sand bed increases the time of impact. As the athlete lands, the mat compresses, extending the duration over which their momentum is reduced to zero. By increasing the time of impact ($t$), the magnitude of the stopping force ($F$) is significantly decreased for the same change in momentum. This smaller force is much safer for the athlete.

This scenario is governed by Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

During the collision, the force exerted by the loaded cart on the empty cart (the 'action') is exactly equal in magnitude and opposite in direction to the force exerted by the empty cart on the loaded cart (the 'reaction').

While the masses of the carts are different and their resulting accelerations will be different (the lighter, empty cart will experience a much larger acceleration), the forces they exert on each other are always equal in magnitude.

Correct statement: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

First, we need to determine the constant force that is being applied. According to Newton's second law of motion: $F = m \times a$

From the given acceleration-mass graph (Fig. 6.40), we can pick any point to calculate the force. Let's choose the point where mass ($m$) = 2 kg. The corresponding acceleration ($a$) is 5 m/s². $F = 2 \text{ kg} \times 5 \text{ m/s}^2 = 10 \text{ N}$

Let's verify with another point. Where mass ($m$) = 5 kg, the acceleration ($a$) is 2 m/s². $F = 5 \text{ kg} \times 2 \text{ m/s}^2 = 10 \text{ N}$

This confirms that the force applied is constant and its value is 10 N.

The question asks to plot the force-mass graph for this case. Since the force is constant at 10 N regardless of the mass, the graph will be a horizontal line parallel to the mass-axis at the value F = 10 N.

(A sketch of the graph would show a horizontal line starting from the y-axis at the '10' mark and extending parallel to the x-axis).

Given: Mass of the object, $m = 10 \text{ kg}$.

To Find: Force acting on the object, $F$.

From the Graph (Fig. 6.41): The graph is a straight line, which indicates constant acceleration. Initial velocity (at $t=0$ s), $u = 20 \text{ m/s}$. Final velocity (at $t=10$ s), $v = 0 \text{ m/s}$. Time interval, $\Delta t = 10 \text{ s} - 0 \text{ s} = 10 \text{ s}$.

Calculation: First, we calculate the acceleration of the object. $a = \frac{0 \text{ m/s} - 20 \text{ m/s}}{10 \text{ s}} = \frac{-20 \text{ m/s}}{10 \text{ s}} = -2 \text{ m/s}^2$

The negative sign indicates that the acceleration is in the direction opposite to the initial velocity (it is a deceleration or retardation).

Now, we calculate the force using Newton's second law. $F = m \times a = 10 \text{ kg} \times (-2 \text{ m/s}^2) = -20 \text{ N}$

Final Answer: The force acting on the object is 20 N. The negative sign signifies that the force is acting in the direction opposite to the motion of the object.

Given: Mass of the bullet, $m = 50 \text{ g} = 0.05 \text{ kg}$. Initial speed, $u = 100 \text{ m/s}$. Final speed, $v = 0 \text{ m/s}$ (since it stops). Distance penetrated, $s = 50 \text{ cm} = 0.5 \text{ m}$.

To Find: Stopping force acting on the bullet, $F$.

Calculation: First, we find the acceleration ($a$) of the bullet inside the block. $v^2 = u^2 + 2as$ $0^2 = (100)^2 + 2 \times a \times 0.5$ $0 = 10000 + a$ $a = -10000 \text{ m/s}^2$

The negative sign shows that the acceleration is opposite to the direction of the bullet's motion (retardation).

Now, we calculate the stopping force. $F = m \times a$ $F = 0.05 \text{ kg} \times (-10000 \text{ m/s}^2)$ $F = -500 \text{ N}$

Final Answer: The estimated stopping force acting on the bullet is 500 N. The negative sign indicates the force opposes the bullet's motion.

Given: Mass of the football, $m = 0.4 \text{ kg}$. Initial velocity of the ball, $u = 0 \text{ m/s}$ (it was stationary for a penalty shot). Final velocity of the ball, $v = 108 \text{ km/h}$. Force imparted, $F = 800 \text{ N}$.

To Find: Time of contact, $t$.

Unit Conversion: First, convert the final velocity to m/s. $v = 108 \frac{\text{km}}{\text{h}} = 108 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 30 \text{ m/s}$

Formula: From Newton's second law of motion, $F = ma$. We also know that $a = \frac{v - u}{t}$. Substituting for $a$, we get: $F = m \frac{(v - u)}{t}$

Calculation: Rearranging the formula to solve for $t$: $t = m \frac{(v - u)}{F}$

Substituting the given values: $t = 0.4 \text{ kg} \times \frac{(30 \text{ m/s} - 0 \text{ m/s})}{800 \text{ N}}$ $t = \frac{0.4 \times 30}{800} \text{ s} = \frac{12}{800} \text{ s} = 0.015 \text{ s}$

Final Answer: The time of contact between the foot and the ball was 0.015 seconds.

Given: Mass of the object, $m = 2 \text{ kg}$. Initial velocity upon entering the patch, $u = 10 \text{ m/s}$. Final velocity, $v = 0 \text{ m/s}$ (it comes to rest). Frictional force, $F_{friction} = 7 \text{ N}$ (opposes motion). Additional opposing force, $F_{applied} = 3 \text{ N}$.

To Find: Distance traveled before coming to rest, $s$.

Final Answer: The object travels 10 meters before coming to rest.

Let the acceleration of the combined system be $a$.

Case 1: Tractor pulls the harrow Force $F$, mass $m_1$, acceleration $a_1$. From Newton's second law: $F = m_1 a_1 \implies m_1 = \frac{F}{a_1}$

Case 2: Tractor pulls the trolley Force $F$, mass $m_2$, acceleration $a_2$. From Newton's second law: $F = m_2 a_2 \implies m_2 = \frac{F}{a_2}$

Case 3: Tractor pulls the harrow and trolley together The total mass of the system is $M = m_1 + m_2$. The force applied is still $F$. Let the new acceleration be $a$. From Newton's second law: $F = M a = (m_1 + m_2)a$

Now, we substitute the expressions for $m_1$ and $m_2$ from Case 1 and Case 2 into the equation for Case 3: $F = \left( \frac{F}{a_1} + \frac{F}{a_2} \right) a$

We can factor out $F$ from the term in the parenthesis: $F = F \left( \frac{1}{a_1} + \frac{1}{a_2} \right) a$

Divide both sides by $F$ (assuming $F \neq 0$): $1 = \left( \frac{1}{a_1} + \frac{1}{a_2} \right) a$

Now, solve for the new acceleration $a$. First, simplify the term in the parenthesis: $1 = \left( \frac{a_2 + a_1}{a_1 a_2} \right) a$

Rearrange to find $a$: $a = \frac{a_1 a_2}{a_1 + a_2}$

Final Answer: The expression for the resulting acceleration is $a = \frac{a_1 a_2}{a_1 + a_2}$.

This observation is explained by combining Newton's second and third laws of motion.

In conclusion, even though the forces are equal, the much smaller mass of the compass needle allows it to have a significant acceleration, while the large mass of the bar magnet results in a negligible acceleration. This is why the needle moves, but the bar magnet does not appear to move.