Kinematics of Particles Questions and Answers

Answer: D

An object moving with negative acceleration has a velocity-time graph sloping downwards to the right, because acceleration is the slope of the velocity-time graph. A displacement-time graph that is a straight line represents a constant velocity, a displacement-time graph curving upwards shows an increasing velocity, and a velocity-time graph sloping upwards shows a positive acceleration.

Negative acceleration, or retardation, means that velocity falls with time, so the line has a negative slope on a velocity-time graph.

Answer: C

The kinetic energy is 16.

In polar coordinates the velocity has a radial part \(\dot{r}\) and a transverse part \(r\dot{\theta}\). At t = 2:

\[ \begin{aligned} \dot{r} &= 2t = 4, \quad r\dot{\theta} = t^{2} \times 1 = 4 \\ KE &= \tfrac{1}{2}(4^{2} + 4^{2}) = 16 \end{aligned} \]

Using only the radial velocity gives 8, and forgetting the factor 1/2 gives 32; neither is offered. 4 and 12 do not follow from these velocity components, and 24 would need one of them to be larger than it is.

Answer: C

The mass comes to rest after 4 s.

A body launched up at speed v returns to the ground after \(2v/g\). The first flight lasts \(2 \times 4/10 = 0.8\) s. Each bounce leaves with 80% of the impact speed, so each flight lasts 0.8 times the one before, and the times form a geometric series:

\[ \begin{aligned} T &= 0.8(1 + 0.8 + 0.8^{2} + \cdots) \\ &= \frac{0.8}{1 - 0.8} = 4\ \text{s} \end{aligned} \]

The number of bounces is infinite, but their durations shrink geometrically, so the total time is finite; that is the trap in the infinity option. 1 s and 2 s stop the series too early (2 s is roughly the first three flights).

Answer: C

The z-component is \(-1.11\) m/s.

The velocity lies in the plane. Split it along the dotted line parallel to AB, which is the line of steepest slope at angle \(\theta\) to the xy-plane, and along the y direction, which is horizontal. Only the AB component has a vertical part, and the particle moves down the slope:

\[ \begin{aligned} v_{z} &= -v\cos\phi\sin\theta \\ &= -2 \times \cos 30^{\circ} \times \sin 40^{\circ} \\ &= -2 \times 0.866 \times 0.643 = -1.11 \end{aligned} \]

\(-1.32\) uses \(\cos 40^{\circ}\) in place of \(\sin 40^{\circ}\), \(-1.00\) is \(-v\sin\phi\), which ignores the slope altogether, and \(-1.50\) does not follow from these angles.

Answer: D

Car P hits car Q at 4 m/s.

Convert the speeds: 36 kmph = 10 m/s and 18 kmph = 5 m/s. In 30 s car Q moves 5 × 30 = 150 m, so car P must cover the 60 m gap plus 150 m, that is 210 m, in the same 30 s with uniform deceleration a:

\[ \begin{aligned} 210 &= 10(30) - \tfrac{1}{2}a(30)^{2} \\ 450a &= 90 \Rightarrow a = 0.2\ \text{m/s}^{2} \\ v &= 10 - 0.2 \times 30 = 4\ \text{m/s} \end{aligned} \]

Check: the average speed of P is (10 + 4)/2 = 7 m/s, and 7 × 30 = 210 m, as required.

5 m/s is car Q's speed, which P would have only if it just touched Q without closing on it. 16 m/s is faster than P's starting speed, impossible under braking, and 1 m/s does not satisfy the distance condition.