JEE Physics Motion In A Plane MCQs Set 17

Answer: (b) 18.03 m/s

Explanation:
1. The racer's velocity is perpendicular to the radius, so it makes 60° with the track: \( \vec v_R=15\cos60^\circ\,\hat i+15\sin60^\circ\,\hat j=7.5\hat i+7.5\sqrt3\,\hat j \).
2. Cyclist: \( \vec v_C=20\hat i \).
3. \( \vec v_{RC}=\vec v_R-\vec v_C=-12.5\hat i+7.5\sqrt3\,\hat j \).
4. \( |\vec v_{RC}|=\sqrt{12.5^2+(7.5\sqrt3)^2}=\sqrt{325}\approx18.03\ \text{m/s} \).

Teacher's Note:
1. Relative velocity is always the vector difference of the two velocities, never the difference of speeds.
2. Velocity on a circle is tangential, perpendicular to the radius.

Answer: (c) The average velocity of the object is zero

Explanation:
1. Initial and final speeds are both zero, so the change in kinetic energy is zero and by the work-energy theorem the work done is zero.
2. Average acceleration = change in velocity / time = 0, so the average force is also zero.
3. The velocity is always positive, so the displacement is the area under the graph, \( \frac12\times30\times20=300\ \text{m} \), and the average velocity is \( 300/30=10\ \text{m/s} \), which is not zero.
4. Hence statement (c) is wrong.

Teacher's Note:
1. Average velocity needs the total displacement; the object never reverses direction here.
2. Zero net change in velocity gives zero average acceleration but not zero displacement.

Answer: (a) \( \frac34 \)

Explanation:
1. Acceleration and retardation phases each last \( \tau=v/a \) and cover \( \tfrac12v\tau \). Let the uniform phase last \( t_2 \).
2. Distance \( =v\tau+vt_2 \), total time \( =2\tau+t_2 \).
3. Average speed \( \frac{v(\tau+t_2)}{2\tau+t_2}=\frac{7v}{8}\Rightarrow 8\tau+8t_2=14\tau+7t_2\Rightarrow t_2=6\tau \).
4. Fraction of time at constant velocity = \( \frac{6\tau}{8\tau}=\frac34 \).

Teacher's Note:
1. The speed-time graph is a trapezoid; its area is the distance.
2. Use the symmetric ramps to reduce the unknowns to tau and t2.

Answer: (d) the velocity of the body at A is zero

Explanation:
1. Velocity is the slope of the displacement-time graph.
2. The curve is not a straight line, so the velocity is not constant and the acceleration is not zero from O to A (the slope keeps decreasing).
3. At A the tangent is horizontal, so the slope is zero, hence the velocity is zero at A.

Teacher's Note:
1. Slope of the s-t graph gives velocity; change of slope gives acceleration.
2. A horizontal tangent means the body is momentarily at rest.

Answer: (c) \( H-\frac12g(t-T)^2 \)

Explanation:
1. At the top the velocity is zero, so measured from the highest point the particle falls freely under gravity.
2. Distance below the top at time t (before or after T): \( \frac12g(t-T)^2 \).
3. Height = \( H-\frac12g(t-T)^2 \).

Teacher's Note:
1. Motion is symmetric about the highest point.
2. Check: at t = T the formula gives H, and at t = 0 it gives 0 because H = gT²/2.

Answer: (b) 32.25 m

Explanation:
1. For 0 to 1.5 s: \( s_1=7\times1.5=10.5\ \text{m} \).
2. For 1.5 s to 3 s (1.5 s duration, initial speed 7 m/s): \( s_2=7(1.5)+\tfrac12(10)(1.5)^2=10.5+11.25=21.75\ \text{m} \).
3. Total \( =10.5+21.75=32.25\ \text{m} \).

Teacher's Note:
1. Split the motion at the instant the acceleration starts.
2. The initial speed of the second stage is still 7 m/s.

Answer: (a) 19.6 m/s

Explanation:
1. Time of flight \( t=\frac{58.8}{14.7}=4\ \text{s} \).
2. The ball must keep the truck's horizontal velocity, so relative to the truck it goes straight up and down.
3. Time of flight \( t=\frac{2u}{g}\Rightarrow u=\frac{gt}{2}=\frac{9.8\times4}{2}=19.6\ \text{m/s} \).

Teacher's Note:
1. Seen from the truck, the horizontal motion disappears.
2. Time of flight for vertical throw is 2u/g.

Answer: (a) 61.9 m

Explanation:
1. Vertical component of initial velocity (downward): \( 25\sin30^\circ=12.5\ \text{m/s} \).
2. Height \( h=12.5(2.5)+\tfrac12(9.8)(2.5)^2=31.25+30.625=61.875\ \text{m} \).

Teacher's Note:
1. Only the vertical component decides how far it falls.
2. Both the initial downward speed and gravity contribute.

Answer: (d) \( R=4\sqrt{hh'} \)

Explanation:
1. The two angles are θ and 90° − θ. \( h=\frac{u^2\sin^2\theta}{2g} \), \( h'=\frac{u^2\cos^2\theta}{2g} \).
2. \( hh'=\frac{u^4\sin^2\theta\cos^2\theta}{4g^2}\Rightarrow\sqrt{hh'}=\frac{u^2\sin\theta\cos\theta}{2g} \).
3. \( R=\frac{u^2\sin2\theta}{g}=\frac{2u^2\sin\theta\cos\theta}{g}=4\sqrt{hh'} \).

Teacher's Note:
1. Complementary angles of projection give equal ranges.
2. Multiply the two heights so that sin and cos combine into sin 2θ.

Answer: (c) decreases by 9%

Explanation:
1. Without air resistance: \( t_0=\frac{u}{g} \).
2. With air resistance the upward deceleration is \( g+0.1g=1.1g \), so \( t=\frac{u}{1.1g}=\frac{t_0}{1.1} \).
3. Change \( =\frac{1}{1.1}-1\approx-0.09 \): a decrease of about 9%.

Teacher's Note:
1. While rising, air drag acts downwards, adding to gravity.
2. Larger deceleration means the particle stops sooner.

Answer: (b) \( \sqrt{\frac{\alpha}{2\beta}} \)

Explanation:
1. No acceleration along x, so \( v_x \) is constant and \( x=v_xt \).
2. Along y (starting with zero y-velocity): \( y=\tfrac12\alpha t^2 \).
3. Substituting in \( y=\beta x^2 \): \( \tfrac12\alpha t^2=\beta v_x^2t^2 \), so \( v_x=\sqrt{\frac{\alpha}{2\beta}} \).

Teacher's Note:
1. Constant acceleration only along y means v_x stays constant.
2. Initial velocity along y is zero because the path y = βx² passes through the origin with zero slope.

Answer: (c) \( \frac{t_1}{t_2}=\tan\theta_2 \)

Explanation:
1. Equal range with equal speed requires \( \sin2\theta_1=\sin2\theta_2 \), so \( \theta_1+\theta_2=90^\circ \) (a is correct).
2. Time of flight \( t=\frac{2u\sin\theta}{g} \), so \( \frac{t_1}{t_2}=\frac{\sin\theta_1}{\sin\theta_2} \) (d is correct).
3. With \( \theta_2=90^\circ-\theta_1 \): \( \frac{t_1}{t_2}=\frac{\sin\theta_1}{\cos\theta_1}=\tan\theta_1 \) (b correct), so (c), which says tan θ2, is incorrect.

Teacher's Note:
1. Complementary angles give the same range.
2. Express sin θ2 as cos θ1 to get the ratio of times.

Answer: (a) \( (2\hat i+4\hat j) \) m/s

Explanation:
1. \( x=\frac{t^2}{2}\Rightarrow v_x=t=2\ \text{m/s} \) at t = 2 s.
2. \( y=\frac{x^2}{2}=\frac{t^4}{8}\Rightarrow v_y=\frac{t^3}{2}=4\ \text{m/s} \).
3. Velocity \( =2\hat i+4\hat j \) m/s.

Teacher's Note:
1. Write y as a function of t before differentiating.
2. Differentiate each coordinate separately.

Answer: (d) 2 m

Explanation:
1. Time of flight \( t=\frac{2u}{g}=\frac{2\times9.8}{9.8}=2\ \text{s} \).
2. The ball keeps the car's initial horizontal velocity, while the car gains extra displacement \( \tfrac12at^2=\tfrac12(1)(2)^2=2\ \text{m} \) relative to the ball.
3. So the ball falls 2 m behind the boy.

Teacher's Note:
1. The ball has no horizontal acceleration after leaving the hand, but the car does.
2. The difference of horizontal displacements is the gap.

Answer: (c) \( 3\hat i \) m

Explanation:
1. Kinetic friction \( =\mu mg=0.1\times4\times10=4\ \text{N} \) acts along −x while the block moves in +x.
2. Deceleration \( a=\frac{2+4}{4}=1.5\ \text{m/s}^2 \), so it stops after \( t=\frac{3}{1.5}=2\ \text{s} \) having moved \( s=\frac{3^2}{2\times1.5}=3\ \text{m} \).
3. Now the applied force 2 N is less than the maximum static friction 4 N, so the block stays at rest for the remaining 3 s.
4. Displacement \( =3\hat i \) m.

Teacher's Note:
1. After stopping, check whether the applied force can overcome limiting friction.
2. Friction can never reverse the motion.

Answer: (b) 3/4 kg

Explanation:
1. The lower block is driven only by friction, so its greatest acceleration is \( a_{max}=\frac{\mu m_1g}{m_2}=\frac{0.1\times2\times10}{1}=2\ \text{m/s}^2 \).
2. For the whole system moving together: \( a=\frac{Mg}{3+M} \).
3. Set \( \frac{10M}{3+M}=2\Rightarrow8M=6\Rightarrow M=\tfrac34\ \text{kg} \).

Teacher's Note:
1. Find the largest acceleration the friction can give the lower block.
2. Then apply Newton's second law to the whole system of three masses.