Chapter 6: How Forces Affect Motion (Physics)

The table moves with constant velocity, so its acceleration is zero and the net force on it is zero (Newton's first law). The only horizontal forces are the applied force F and friction, so they must balance:

$$F-f=0\Rightarrow f=F$$

The frictional force is equal to F in magnitude, acting opposite to the applied force.

(i) Remain the same. With no net force there is no acceleration (first law), so the ball keeps moving with the same velocity.

(ii) Increase. The force produces an acceleration in the direction of motion, so the ball speeds up.

(iii) Decrease. The acceleration is opposite to the motion (retardation), so the ball slows down.

(i) remain the same (ii) increase (iii) decrease

Answer: (i)

• On P the forces are unbalanced: net force $=5-4=1$ N, in the direction of the 5 N force. So P experiences a net force (and accelerates).
• Q moves with constant velocity, so its acceleration is zero and the net force on it is zero.

(i)

95 oarsmen push the boat forward and 5 push it backward:

$$\text{Forward force}=95\times 200=19000\text{N},\text{Backward force}=5\times 200=1000\text{N}$$

$$\text{Net force}=19000-1000=18000\text{N}$$

18,000 N in the forward direction.

Answer: (iv)

By Newton's second law, $a=\frac{F}{m}$: the acceleration is in the direction of the net force and is directly proportional to the force (and inversely proportional to the mass).

(iv)

Answer: (iii) Object C

• A and D: straight sloping lines, so constant velocity (A moving forward, D moving back). No acceleration, so no net force.
• B: horizontal line; position does not change, so B is at rest. No net force.
• C: a curve that gets steeper with time; its velocity (slope) keeps increasing, so C is accelerating. A net force must act on it.

(iii) Object C, the only object whose velocity changes (curved graph).

Yes, the boat moves backward (away from the shore).

To jump forward, the sailor pushes the boat backward with their feet. By Newton's third law, the boat pushes the sailor forward with an equal and opposite force. The force on the boat (backward) makes it move backward. Since the boat is light and floats on water (little friction), it moves back easily. This also follows from conservation of momentum: before the jump the total momentum is zero, so the forward momentum of the sailor is balanced by an equal backward momentum of the boat.

Yes. The boat moves backward, because the sailor pushes the boat back and the boat pushes the sailor forward (equal and opposite action and reaction).

When the athlete lands, their momentum must be reduced to zero. The force on the body equals the rate of change of momentum:

$$F=\frac{\text{change in momentum}}{\text{time taken}}$$

The change in momentum is the same whether they land on hard ground or on a mat. A soft mat or sand bed gets pressed in, so the athlete takes a longer time to stop. A longer time means a smaller force on the body, so the athlete is not injured. On hard ground they would stop almost instantly, and the large force could hurt them.

The mat increases the time taken to stop the athlete, so the rate of change of momentum, and hence the force on the body, becomes small and prevents injury.

Answer: (iv)

By Newton's third law, the forces the two carts exert on each other are an action–reaction pair: equal in magnitude and opposite in direction, whatever their masses. (Their effects differ: the lighter empty cart gets a larger acceleration.)

(iv)

From Fig. 6.40, read some points and find $F=ma$ for each:

The same force, 10 N, acts in every case (that is why $a$ falls as $m$ rises). So the force–mass graph is a straight line parallel to the mass axis at $F=10$ N.

F = ma = 10 N for every point, so the force–mass graph is a horizontal straight line at F = 10 N.

From Fig. 6.41, the graph is a straight line: the velocity goes from 10 m s⁻¹ at t = 0 to 30 m s⁻¹ at t = 8 s (it is 20 m s⁻¹ at 4 s).

$$a=\text{slope}=\frac{30-10}{8-0}=2.5{\text{m s}}^{-2}$$

$$F=ma=10\times 2.5=25\text{N}$$

25 N (in the direction of motion)

$m=50$ g $=0.05$ kg, $u=100$ m s⁻¹, $v=0$, $s=50$ cm $=0.5$ m.

$$v^2=u^2+2as\Rightarrow 0=10000+2a(0.5)\Rightarrow a=-10000{\text{m s}}^{-2}$$

$$F=ma=0.05\times (-10000)=-500\text{N}$$

500 N, opposite to the motion of the bullet.

$u=0$, $v=108\times \frac{5}{18}=30$ m s⁻¹, $m=0.4$ kg, $F=800$ N.

$$F=\frac{m(v-u)}{t}\Rightarrow t=\frac{m(v-u)}{F}=\frac{0.4\times 30}{800}=\frac{12}{800}=0.015\text{s}$$

0.015 s (15 milliseconds)

Both forces oppose the motion, so the net retarding force $=7+3=10$ N.

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

With $u=10$ m s⁻¹ and $v=0$:

$$v^2=u^2+2as\Rightarrow 0=100-10s\Rightarrow s=10\text{m}$$

10 m

From $F=ma$:

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

With the harrow on the trolley, the total mass is $m_1+m_2$, so the new acceleration is

$$a=\frac{F}{m_1+m_2}=\frac{F}{\frac{F}{a_1}+\frac{F}{a_2}}=\frac{1}{\frac{1}{a_1}+\frac{1}{a_2}}=\frac{a_1{a}_2}{a_1+a_2}$$

$a=\frac{a_1{a}_2}{a_1+a_2}$

The two forces are equal, but they act on different objects, and the effect of a force depends on the mass of the object and the other forces acting on it ($a=\frac{F}{m}$):

• The compass needle is very light and is balanced on a pivot with almost no friction, so even a small force gives it a large (turning) acceleration: it moves.
• The bar magnet is much heavier and is held in the hand (or rests on the table, where friction acts). The small magnetic force is balanced by the hand or friction, and in any case it would give the heavy magnet only a negligible acceleration, so it does not appear to move.

The forces are equal, but the light, nearly frictionless compass needle gets a large acceleration, while the heavier bar magnet, held in the hand or by friction, gets almost none.