How Forces Affect Motion — NCERT Solutions

For block P, the forces are 4 N and 5 N in opposite directions, so the net force is 1 N. Hence P experiences a net force.

Block Q is moving with constant velocity, so its net force is zero.

Therefore, the correct statement is (i).

There are 95 oarsmen rowing one way and 5 rowing the opposite way.

Net effective oarsmen = $95 - 5 = 90$

Each exerts $200\,\text{N}$, so
$$\text{Net force} = 90 \times 200 = 18000\,\text{N}$$
So the net force on the snake boat is 18000 N forward.

By Newton’s second law, acceleration is in the direction of force and is proportional to the force.

So the correct option is (iv).

A net force acts when an object has acceleration. On a position-time graph, acceleration is indicated by a curved graph rather than a straight line.

From Fig. 6.37, the object with a curved position-time graph is Object A.

Yes, the boat will move backward. When the sailor jumps forward, the sailor exerts a force on the boat, and the boat exerts an equal and opposite force on the sailor. Since the forces act on different objects, they do not cancel. The boat recoils backward due to Newton’s third law of motion.

The landing mat or sand bed increases the time over which the athlete comes to rest. For the same change in velocity, a larger time means smaller acceleration, and therefore a smaller force on the body. This reduces injury and makes landing safer.

By Newton’s third law, the loaded cart and the empty cart exert equal and opposite forces on each other during collision. So the correct option is (iv).

A force-mass graph for this case will be a horizontal straight line.

Reason: from Newton’s second law,
$$F = ma$$
If the same force produces different accelerations on different masses, then for this case the force remains constant while mass changes. So when force is plotted against mass, the graph is parallel to the mass axis.

From the graph, the car’s velocity changes from $0$ to $10\,\text{m s}^{-1}$ in $5\,\text{s}$.

So acceleration is
$$a=\frac{v-u}{t}=\frac{10-0}{5}=2\,\text{m s}^{-2}$$
Using Newton’s second law,
$$F=ma=10\times 2=20\,\text{N}$$
But the chapter’s example with the same graph and mass shows the force as 3000 N east for a $1500\,\text{kg}$ car. Since this question states the mass is 10 kg, the force is
$$F=10\times 2=20\,\text{N}$$
So the computed answer from the given mass is 20 N in the direction of motion. The printed example value of 3000 N does not match this mass.

Given:

• mass of bullet, $m=50\,\text{g}=0.05\,\text{kg}$
• initial speed, $u=100\,\text{m s}^{-1}$
• final speed, $v=0$
• stopping distance, $s=50\,\text{cm}=0.5\,\text{m}$

Using
$$v^2=u^2+2as$$
$$0=(100)^2+2a(0.5)$$
$$0=10000+a$$
$$a=-10000\,\text{m s}^{-2}$$
Now,
$$F=ma=0.05\times(-10000)=-500\,\text{N}$$
So the stopping force has magnitude 500 N, opposite to the motion.

However, if the intended estimate is from work-energy with the same numbers:
$$F\times 0.5 = \tfrac12 \times 0.05 \times 100^2 = 250 \Rightarrow F=500\,\text{N}$$
So the correct stopping force is 500 N.

Given:

• force, $F=800\,\text{N}$
• mass, $m=0.4\,\text{kg}$
• initial speed of football, $u=108\,\text{km h}^{-1}=30\,\text{m s}^{-1}$
• final speed after kick starts from rest at contact, so use change in momentum during contact

Using Newton’s second law,
$$a=\frac{F}{m}=\frac{800}{0.4}=2000\,\text{m s}^{-2}$$
Now,
$$v=u+at$$
Here the ball goes from rest to $30\,\text{m s}^{-1}$, so
$$30=0+2000t$$
$$t=\frac{30}{2000}=0.015\,\text{s}$$
So the time of contact is 0.015 s.

The backward forces are friction $7\,\text{N}$ and the additional opposing force $3\,\text{N}$.

So, total retarding force:

$$F = 7 + 3 = 10\,\text{N}$$

Mass of the object:

$$m = 2\,\text{kg}$$

Using Newton's second law,

$$a = \frac{F}{m} = \frac{10}{2} = 5\,\text{m s}^{-2}$$

This acceleration is opposite to the motion, so take $a = -5\,\text{m s}^{-2}$.

Initial velocity $u = 10\,\text{m s}^{-1}$ and final velocity $v = 0$.

Using

$$v^2 = u^2 + 2as$$

$$0 = 10^2 + 2(-5)s$$

$$0 = 100 - 10s$$

$$10s = 100$$

$$s = 10\,\text{m}$$

However, this result is not consistent with the usual interpretation here because the object stops under a constant opposing force of $10\,\text{N}$ over the rough patch. The correct calculation is:

$$a = \frac{10}{2} = 5\,\text{m s}^{-2}$$

$$0 = 100 - 2\times 5 \times s$$

$$0 = 100 - 10s$$

$$s = 10\,\text{m}$$

So the object travels 10 m before coming to rest.

For the harrow of mass $m_1$,

$$F = m_1 a_1$$

so

$$m_1 = \frac{F}{a_1}$$

For the trolley of mass $m_2$,

$$F = m_2 a_2$$

so

$$m_2 = \frac{F}{a_2}$$

When the tractor pulls the trolley with the harrow placed on it, the total mass is

$$m_1 + m_2 = \frac{F}{a_1} + \frac{F}{a_2}$$

The same force $F$ acts on the combined system, so the resulting acceleration $a$ is

$$a = \frac{F}{m_1+m_2}$$

Substitute the masses:

$$a = \frac{F}{\frac{F}{a_1}+\frac{F}{a_2}}$$

$$a = \frac{1}{\frac{1}{a_1}+\frac{1}{a_2}}$$

$$a = \frac{a_1a_2}{a_1+a_2}$$

The bar magnet and the compass needle exert equal and opposite forces on each other, but they do not necessarily produce the same effect because the two objects have different masses.

By Newton's second law,

$$a = \frac{F}{m}$$

So, for the same force:

• the compass needle, having a much smaller mass, gets a larger acceleration and moves easily;
• the bar magnet, having a much larger mass, gets a very small acceleration, so its motion is not noticed.

Thus, the forces are equal and opposite, but the bar magnet does not seem to move because its acceleration is extremely small.