Laws of Motion Questions and Answers

Answer: B

The condition is \(v^{2} \leq \mu R g\).

On a level road the only horizontal force on the cycle is static friction, and it must supply the centripetal force for the turn. The largest friction available is \(\mu m g\):

\[ \begin{aligned} \frac{m v^{2}}{R} &\leq \mu m g \\ v^{2} &\leq \mu R g \end{aligned} \]

The mass cancels. The inequality must run this way, since a higher speed needs more friction; options with \(\geq\) describe the condition for slipping. \(\mu R\) and \(\frac{\mu R}{g}\) do not have the units of \(v^{2}\), and \(v \leq \mu R g\) leaves \(v\) unsquared.

Answer: A

The impulse is 4.5 kg m s^-1.

54 km/h = 15 m/s. The ball reverses with the same speed, so its momentum changes from \(+mv\) to \(-mv\):

\[ \begin{aligned} J &= 2mv = 2(0.15)(15) \\ &= 4.5\ \text{kg m s}^{-1} \end{aligned} \]

16.2 is \(2 \times 0.15 \times 54\), with the speed left in km/h, and 8.1 is \(0.15 \times 54\) without the factor 2 as well. The reversal is what doubles the change in momentum.

Answer: B

The mass of the body is 2.5 kg.

Perpendicular forces combine by Pythagoras, and Newton's second law then gives the mass.

\[ \begin{aligned} F &= \sqrt{3^{2} + 4^{2}} = 5 \text{ N} \\ m &= \frac{F}{a} = \frac{5}{2} = 2.5 \text{ kg} \end{aligned} \]

3.5 kg comes from adding the forces directly (7 N), which is wrong for perpendicular forces, and 1.5 kg or 2 kg use only one of them.

Answer: C

The incorrect statement is Newton's second law is inconsistent with Newton's first law.

The second law, \(\vec{F} = \frac{d\vec{p}}{dt}\), contains the first: when the net external force is zero, momentum does not change, so a body stays at rest or in uniform motion. The two laws are consistent.

The other statements are correct. The second law relates net external force to acceleration. Impulse equals change in momentum. Since \(\Delta p = F\Delta t\), the same force for the same time gives the same change in momentum whatever the body's mass. Action and reaction act on different bodies, which is why they do not cancel.

Answer: A

The contact time is 2 s.

The ball reverses with the same speed, so its momentum changes from \(+mv\) to \(-mv\):

\[ \begin{aligned} \Delta p &= 2mv = 2(0.2)(5) = 2\ \text{N s} \\ t &= \frac{\Delta p}{F} = \frac{2}{1} = 2\ \text{s} \end{aligned} \]

1 s is the answer you get by taking the change in momentum as \(mv\), forgetting that the ball bounces back. 0.5 s halves it again.

Answer: B

The recoil force is 5 N.

The force on the gun equals the rate at which it gives momentum to the bullets:

\[ \begin{aligned} F &= n \cdot m v \\ &= 25 \times 0.01 \times 20 = 5\ \text{N} \end{aligned} \]

The mass must be in kilograms (10 g = 0.01 kg). 50 N comes from using 0.1 kg, and 10 N or 20 N from doubling or squaring a factor.

Answer: E

The minimum speed at the bottom is \(\sqrt{5gr}\).

To complete the circle the string must stay taut at the top, so at the top gravity alone can supply the centripetal force: \(v_{top}^{2} = gr\). Energy conservation from bottom to top (height gain 2r) then gives the bottom speed.

\[ \begin{aligned} v_{bottom}^{2} &= v_{top}^{2} + 2g(2r) \\ &= gr + 4gr = 5gr \end{aligned} \]

\(\sqrt{gr}\) is the minimum at the top, not the bottom. \(\sqrt{2gr}\) only lifts the particle to the level of the centre, and \(\sqrt{3gr}\) is the minimum speed at that horizontal position.

Answer: B

The impulse is 3.6 N s.

Impulse equals the change in momentum. The velocity reverses, so take the rebound direction as positive: the velocity changes from -10 m/s to +8 m/s.

\[ \begin{aligned} J &= m(v - u) \\ &= 0.2 \times (8 - (-10)) \\ &= 0.2 \times 18 = 3.6\ \text{N s} \end{aligned} \]

Ignoring the reversal gives \(0.2 \times (10 - 8) = 0.4\) N s, and 1.8 N s is half the correct value. The impulse on the wall has the same size, 3.6 N s.

Answer: A

The velocity is \(\dfrac{kt^{2}}{2m}\).

The acceleration is \(a = \frac{F}{m} = \frac{kt}{m}\), which changes with time, so integrate it:

\[ v = \int_{0}^{t} \frac{kt}{m}\,dt = \frac{kt^{2}}{2m} \]

Equivalently, the impulse \(\int F\,dt = \frac{kt^{2}}{2}\) equals the momentum gained. \(\frac{kt^{2}}{m}\) forgets the \(\frac{1}{2}\) from integration (it treats the force as constant at its final value), and the options with \(k^{2}\) are dimensionally wrong.

Answer: A

The scale reads \(\dfrac{3mg}{2}\).

The scale reads the normal force N on the passenger. Taking upward as positive, Newton's second law gives

\[ \begin{aligned} N - mg &= m\frac{g}{2} \\ N &= \frac{3mg}{2} \end{aligned} \]

The passenger feels heavier when the lift accelerates upward. \(\frac{mg}{2}\) is the reading for a downward acceleration of \(\frac{g}{2}\), and \(mg\) (also written \(\frac{2mg}{2}\)) is the reading at rest or at constant velocity.