Laws of Motion Questions and Answers

Answer: C

The tension in each cable is 98 N.

The vertical components of the two equal tensions support the weight:

\[ \begin{aligned} 2T\cos 30^{\circ} &= mg \\ T &= \frac{10\sqrt{3} \times 9.8}{2 \times \frac{\sqrt{3}}{2}} = 98\ \text{N} \end{aligned} \]

The horizontal components cancel. 9.8 N and 19.6 N drop the factor 10 in the mass, and 10 N and 20 N use the mass as if it were a weight.

Answer: E

The final velocity is 260 ms^-1.

The force acts along the motion, so it adds to the speed. By Newton's second law the acceleration is

\[ \begin{aligned} a &= \frac{F}{m} = \frac{50}{2} = 25\ \text{m s}^{-2} \\ v &= u + at = 10 + 25 \times 10 = 260\ \text{m s}^{-1} \end{aligned} \]

The impulse-momentum form gives the same: Ft = 500 N s is the change in momentum, which divided by 2 kg is 250 m/s added to 10 m/s. 250 (not offered) forgets the initial velocity; 200 and 220 come from slips in the arithmetic.

Answer: D

The force on the wall is \(1 \times 10^{4}\) N.

The body reverses direction, so the change in its momentum is twice its momentum, and the average force is that change divided by the contact time:

\[ \begin{aligned} \Delta p &= m(v - (-v)) = 5 \times 100 = 500\ \text{kg m s}^{-1} \\ F &= \frac{\Delta p}{\Delta t} = \frac{500}{1/20} = 10^{4}\ \text{N} \end{aligned} \]

\(0.5 \times 10^{4}\) N uses only \(mv = 250\), forgetting that the rebound doubles the change in momentum. \(2.5 \times 10^{4}\) N and the \(10^{3}\) values come from mishandling the \(\frac{1}{20}\) s.

Answer: E

The force is 5 N.

The force on the gun equals the momentum given to the bullets per second:

\[ \begin{aligned} F &= n\,m\,v \\ &= \frac{25}{60}\ \text{s}^{-1} \times 0.04\ \text{kg} \\ &\quad \times 300\ \text{m/s} \\ &= \frac{25 \times 12}{60} = 5\ \text{N} \end{aligned} \]

The gun's own mass of 5 kg does not enter. Using 25 bullets per second instead of per minute gives 300 N, and forgetting to convert grams gives 5000 N; neither is offered, while 2.5 N, 4 N, 7 N and 10 N do not follow from the data.

Answer: D

The mass is 1.5 kg.

The acceleration given is the magnitude, so use the magnitude of the force in \(F = ma\):

\[ \begin{aligned} |\vec{F}| &= \sqrt{1^{2} + 2^{2} + (-2)^{2}} = 3\ \text{N} \\ m &= \frac{|\vec{F}|}{a} = \frac{3}{2} = 1.5\ \text{kg} \end{aligned} \]

Adding the components instead (\(1 + 2 - 2 = 1\)) gives 0.5 kg. 5 kg, 7 kg and 10 kg do not follow from a 3 N force and a 2 m s^-2 acceleration.

Answer: A

The average force is 75 N.

Impulse is average force times the time of contact, \(J = F\Delta t\):

\[ F = \frac{J}{\Delta t} = \frac{15}{0.2} = 75\ \text{N} \]

15 N treats the impulse as the force, and 50 N or 25 N come from slips in dividing by 0.2.

Impulse equals the change in momentum, so the ball's momentum changes by 15 kg m s^-1 whatever its mass; a shorter contact time would mean a larger force for the same impulse.

Answer: E

The force is 2.8 N.

\[ \begin{aligned} v &= \frac{dx}{dt} = 3 + 14t \\ a &= \frac{dv}{dt} = 14\ \text{m s}^{-2} \\ F &= ma = 0.2 \times 14 = 2.8\ \text{N} \end{aligned} \]

The acceleration is constant, so the force is too. Taking the acceleration as 7 (the coefficient of \(t^{2}\)) gives 1.4 N, not offered; the other values do not follow from \(F = ma\).

The 3t term is a constant initial velocity (3 m s^-1) and needs no force to maintain it.

Answer: A

The force is 20 N.

Newton's second law states that force is the rate of change of momentum:

\[ \begin{aligned} F &= \frac{dp}{dt} = 6t + 2 \\ F(3) &= 18 + 2 = 20\ \text{N} \end{aligned} \]

The constant 1 in \(p\) disappears on differentiating. Putting \(t = 3\) into \(p\) itself gives 34 kg m s^-1, a momentum, not a force; 8 N is the value at \(t = 1\).

Answer: E

The maximum speed is 6 ms^-1.

On a flat circular road, friction supplies the centripetal force. The car can go no faster than the speed at which the needed force equals the maximum friction:

\[ \begin{aligned} \frac{mv^{2}}{r} &= \mu mg \\ v &= \sqrt{\mu gr} = \sqrt{0.3 \times 10 \times 12} \\ &= \sqrt{36} = 6\ \text{m s}^{-1} \end{aligned} \]

The option 36 is \(v^{2}\), forgetting the square root, and 3.6 shifts the decimal point. The mass cancels, so the answer does not depend on the car.

Answer: E

The false statement is there is cause-effect relation between action and reaction.

Newton's third law says forces always occur in pairs: if body B exerts a force on body A, then A exerts an equal and opposite force on B. The two forces arise together, at the same instant, from the same interaction. Neither causes the other, so there is no cause-and-effect relation between them.

The other statements are all true. Forces occur between pairs of bodies; the force on B by A is the negative of the force on A by B; action and reaction are simultaneous; and since neither comes first, either one may be called the action and the other the reaction.