Laws of Motion Questions and Answers

Answer: C

The person feels a weight loss when the lift moves downward with constant acceleration.

The apparent weight is the normal reaction R from the floor. With a downward acceleration a, Newton's second law gives mg − R = ma, so R = m(g − a), less than mg.

At uniform velocity (up or down) a = 0 and R = mg, so there is no change. With an upward acceleration R = m(g + a), an apparent gain in weight.

Answer: A

The acceleration is \(\frac{F_{3}}{m}\).

The body is in equilibrium under the three forces, so \(\vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} = 0\): the resultant of \(F_{1}\) and \(F_{2}\) is equal and opposite to \(F_{3}\). When \(F_{3}\) is removed, the net force is that resultant, of magnitude \(F_{3}\), so the acceleration is \(F_{3}/m\). Since \(F_{1}\) and \(F_{2}\) are perpendicular, this also equals \(\sqrt{F_{1}^{2} + F_{2}^{2}}/m\).

\(F_{1}F_{2}/m\) is dimensionally wrong, and \((F_{1} - F_{2})/m\), \(F_{1}/m\) and \(F_{2}/m\) ignore the vector addition of the two perpendicular forces.

Answer: D

The ratio of accelerations is 3 : 2.

By Newton's second law \(a = F/m\). With the same force on both bodies, acceleration is inversely proportional to mass:

\[ \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}} = \frac{3}{2} \]

The lighter 2 kg body gets the larger acceleration. 2 : 3 is the ratio of the masses, the answer you get by treating acceleration as proportional to mass instead of inversely proportional. 1 : 1 would be true only for equal masses, and 4 : 9 squares the mass ratio for no reason.

Answer: B

The area under a force-time graph is the impulse.

Impulse is defined as force multiplied by the time for which it acts, or for a varying force \(J = \int F\,dt\), which is exactly the area between the curve and the time axis. By Newton's second law, this area also equals the change in momentum of the body.

Work done is the area under a force-displacement graph, not force-time. Power is the rate of doing work, kinetic energy is \(\tfrac{1}{2}mv^{2}\), and torque is force times a perpendicular distance; none of these is a product of force and time.

Answer: C

The ball was bowled at 50 ms⁻¹.

The ball reverses with the same speed \(v\), so its momentum changes from \(-mv\) to \(+mv\). The impulse equals that change:

\[ \begin{aligned} J &= 2mv \\ v &= \frac{2}{2 \times 0.02} = 50\ \text{m s}^{-1} \end{aligned} \]

100 m/s, from forgetting the reversal (\(J = mv\)), is not offered; 20, 40, 60 and 80 m/s do not satisfy \(2mv = 2\).

Answer: C

The correct statement is that \(f_{s}\) and \(f_{k}\) are independent of A.

The laws of friction say that friction depends on the normal force and the nature of the surfaces, not on the apparent area of contact: a larger area spreads the same load more thinly, and the real contact area stays proportional to the load. So both coefficients are independent of \(A\).

The kinetic coefficient is normally smaller than the maximum static one, not greater, and it is static friction, not kinetic, that opposes impending motion; kinetic friction opposes actual sliding.

Answer: C

The mass of the block is 8 kg.

In a lift accelerating upward, the string must both support the weight and supply the upward acceleration: \(T - mg = ma\). Taking g = 9.8 m s\(^{-2}\):

\[ m = \frac{T}{g + a} = \frac{80}{9.8 + 0.2} = 8\ \text{kg} \]

The numbers are chosen so that g + a = 10. Dividing by g - a (lift accelerating down) would give about 8.3 kg, and the smaller options ignore the weight term.

Answer: E

The force is 1 N.

By Newton's second law, force equals mass times the rate of change of velocity. Convert 200 g to 0.2 kg.

\[ \begin{aligned} F &= m\frac{\Delta v}{\Delta t} \\ &= 0.2 \times \frac{25}{5} = 1\ \text{N} \end{aligned} \]

Keeping the mass in grams gives 1000 times too much, and forgetting to divide by the time gives 5 N. 2.5 N and 50 N divide or multiply by the wrong quantity.

Answer: B

The coefficient of friction is the ratio of frictional force to normal force.

\(\mu = \frac{f}{N}\), where f is the limiting (or kinetic) friction and N the normal reaction between the surfaces. It depends on the nature of the surfaces, not on the applied force.

The ratios with the applied force or the weight are wrong (the weight equals N only on a horizontal surface with no other vertical forces), and the inverted ratios are reciprocals of \(\mu\).

Answer: E

The force is 375 N.

Take the final direction as positive. The change in momentum is

\[ \begin{aligned} \Delta p &= 0.15 \times 30 - 0.15 \times (-20) = 7.5\ \text{kg m/s} \\ F &= \frac{\Delta p}{\Delta t} = \frac{7.5}{0.02} = 375\ \text{N} \end{aligned} \]

Ignoring the reversal (30 − 20 = 10 m/s) gives 75 N, not offered; 150 N and 15 N come from using the mass in grams or dropping a factor of ten, and 3.75 N forgets to divide by the contact time.

The force is large because the contact time is short; this is the impulse-momentum theorem, \(F\,\Delta t = \Delta p\).