JEE Physics Motion In A Straight Line MCQs Set 14

Answer: (b) \(\sqrt{\dfrac{2}{3}}\)

Explanation:
1. While rising, both gravity and the resistance act downwards: \(a_{up}=10+2=12\) m/s²
2. While falling, resistance acts upwards: \(a_{down}=10-2=8\) m/s²
3. The same height h is covered both ways, \(h=\tfrac12 a t^2\), so \(t\propto 1/\sqrt a\)
4. \(\dfrac{t_{up}}{t_{down}}=\sqrt{\dfrac{8}{12}}=\sqrt{\dfrac23}\)

Teacher's Note:
1. Air resistance always opposes the motion, so it helps gravity on the way up and opposes it on the way down.
2. Equal distances covered with a larger acceleration take less time.

Answer: (b) 65 m/s

Explanation:
1. The stone returns to the tower top with speed 20 m/s downwards, and the energy balance for the whole motion is the same as for a throw from the top with speed 20 m/s
2. \(v^2=u^2+2gh=400+2(9.8)(200)=4320\)
3. \(v=\sqrt{4320}\approx 65.7\) m/s, which is closest to 65 m/s

Teacher's Note:
1. The final speed depends only on the initial speed and the height, not on the direction of the throw.
2. Use \(v^2=u^2+2gh\) directly instead of splitting the motion into up and down parts.

Answer: (c) 5.4 s

Explanation:
1. The body leaves with the balloon's velocity, so \(u=+12\) m/s (upwards) and the displacement is \(-81\) m
2. \(-81=12t-5t^2\), i.e. \(5t^2-12t-81=0\)
3. \(t=\dfrac{12+\sqrt{144+1620}}{10}=\dfrac{12+42}{10}=5.4\) s

Teacher's Note:
1. An object dropped from a moving carrier keeps the carrier's velocity at the instant of release.
2. Take the upward direction as positive and the final displacement as negative.

Answer: (b) \(\dfrac{4u^2}{g}\)

Explanation:
1. The stone passes the top again moving downwards with speed u, and the final speed is 3u
2. \(v^2=u^2+2gh\Rightarrow 9u^2=u^2+2gh\)
3. \(h=\dfrac{8u^2}{2g}=\dfrac{4u^2}{g}\)

Teacher's Note:
1. Use the speed-height relation \(v^2=u^2+2gh\) for the whole path.
2. Direction of the throw does not matter for the final speed.

Answer: (c) 5 m/s

Explanation:
1. For the dropped body \(3^2=2gh\)
2. For the second body \(v^2=4^2+2gh=16+9=25\)
3. \(v=5\) m/s

Teacher's Note:
1. Both bodies fall through the same height, so \(2gh\) is the same for both.
2. Add the squares of the speeds, not the speeds themselves.

Answer: (b) \(v_A=v_B\)

Explanation:
1. Ball A returns to the launch level with the same speed downwards, so from there it moves exactly like B
2. Hence \(v^2=u^2+2gh\) is the same for both and \(v_A=v_B\)

Teacher's Note:
1. The final speed is fixed by the initial speed and height only.
2. The upward ball takes longer, but arrives with the same speed.

Answer: (c) 600 m

Explanation:
1. The stone starts with the balloon's velocity, 10 m/s downwards
2. \(s=ut+\tfrac12gt^2=10(10)+\tfrac12(10)(100)=100+500\)
3. \(s=600\) m

Teacher's Note:
1. A stone released from a moving balloon shares the balloon's velocity at that instant.
2. Here the initial velocity and g point the same way, so they add.

Answer: (a) 293 m

Explanation:
1. Speed after the free fall: \(v^2=2gh=2(9.8)(50)=980\)
2. Under deceleration: \(3^2=980-2(2)s\Rightarrow s=\dfrac{971}{4}\approx 242.75\) m
3. Total height \(=50+242.75\approx 293\) m

Teacher's Note:
1. Split the motion into the free-fall stage and the retarded stage; the end speed of the first is the start speed of the second.
2. Use \(v^2=u^2+2as\) in each stage.

Answer: (a) 90 s

Explanation:
1. Speed at burn-out \(v=at=20(30)=600\) m/s
2. After burn-out only gravity acts, so the rise continues for \(t'=\dfrac{600}{10}=60\) s
3. Total time \(=30+60=90\) s

Teacher's Note:
1. After the fuel ends the rocket keeps moving up while slowing at g.
2. The maximum height is reached when the velocity becomes zero.

Answer: (c) Area A + Area B + Area C + Area D

Explanation:
1. Distance is the total path length, so every area counts positively whether the velocity is positive or negative
2. Up to \(t_4\) the areas A, B, C and D lie in the interval, while E lies after \(t_4\)
3. Distance \(=\) A + B + C + D

Teacher's Note:
1. Areas below the time axis give negative displacement but positive distance.
2. Read the time limit carefully, since area E lies beyond \(t_4\).

Answer: (a) Change in velocity

Explanation:
1. Area under an acceleration-time graph is \(\int a\,dt\)
2. Since \(a=\dfrac{dv}{dt}\), this integral equals \(\Delta v\)
3. The shaded area therefore represents the change in velocity (change in momentum would need the mass as well)

Teacher's Note:
1. The area under a-t gives change in velocity; the area under v-t gives displacement.
2. Momentum change would need the mass, which is not given.

Answer: (b) B

Explanation:
1. Velocity is the slope of the displacement-time graph
2. At A and C the graph has sharp corners, so the slope changes abruptly and is not zero on either side
3. At B the curve is smooth with a minimum, so the slope is zero and \(v=0\)

Teacher's Note:
1. Zero slope on a smooth displacement-time curve means zero velocity.
2. A sharp peak is a sudden reversal with non-zero velocity on both sides.

Answer: (c) C

Explanation:
1. Instantaneous velocity is the slope of the tangent
2. At C the curve is rising, so the slope is positive
3. At D the slope is zero, at E it is falling (negative), and at F the slope is zero at the minimum

Teacher's Note:
1. Rising curve means positive velocity and falling curve means negative velocity.
2. A turning point of the curve has zero slope.

Answer: (c) 8 m/s

Explanation:
1. Distance covered: 0 to 40 m (40 m), then 40 m to -40 m (80 m), then -40 m back to 0 (40 m), total 160 m
2. Total time is 20 s
3. Average speed \(=\dfrac{160}{20}=8\) m/s

Teacher's Note:
1. Average speed uses the total path length, not the net displacement.
2. The net displacement is zero here, so average velocity would be zero.

Answer: (b) 4 s

Explanation:
1. Speed is the magnitude of velocity
2. The graph crosses the time axis (v = 0) at t = 4 s
3. The speed is zero there, which is its minimum

Teacher's Note:
1. Speed is minimum where the velocity curve crosses zero.
2. At 2 s and 6 s the speed is maximum (10 m/s).

Answer: (d) 1 : 3

Explanation:
1. Velocity equals the slope of the displacement-time graph
2. \(v_A=\tan30^\circ=\dfrac1{\sqrt3}\) and \(v_B=\tan60^\circ=\sqrt3\)
3. \(v_A:v_B=\dfrac{1}{\sqrt3}:\sqrt3=1:3\)

Teacher's Note:
1. The slope is the tangent of the angle made with the time axis, not the angle itself.
2. Equal scales on both axes are assumed.

Answer: (b)

Explanation:
1. In one-dimensional motion a particle has exactly one velocity at any instant
2. The closed loop in (b) gives two velocities for the same time, so it cannot describe one-dimensional motion
3. The other graphs are single-valued functions of time

Teacher's Note:
1. A valid v-t graph must give one value of v for each value of t.
2. A closed curve fails this test.

Answer: (d)

Explanation:
1. Uniform acceleration means constant slope of the v-t graph, with the slope non-zero
2. A straight inclined line through the origin has constant slope
3. (a) has zero slope, and (b) and (c) have changing slopes

Teacher's Note:
1. Constant slope on a v-t graph means constant acceleration.
2. A horizontal line means constant velocity (zero acceleration).

Answer: (d) 6 m

Explanation:
1. 0 to 1 s: \(\tfrac12(1)(4)=2\) m; 1 to 3 s: \(2\times4=8\) m; 3 to 4 s: \(\tfrac12(1)(4)=2\) m
2. 4 to 5 s: \(-\tfrac12(1)(2)=-1\) m; 5 to 7 s: \(-2\times2=-4\) m; 7 to 8 s: \(-\tfrac12(1)(2)=-1\) m
3. Net displacement \(=2+8+2-1-4-1=6\) m

Teacher's Note:
1. Displacement is the signed area under the v-t graph.
2. The particle started from the origin, so displacement equals distance from the origin here.

Answer: (a) 8 m, 16 m

Explanation:
1. 0 to 2 s: \(4\times2=8\) m forward; 2 to 4 s: \(-2\times2=-4\) m; 4 to 6 s: \(2\times2=4\) m
2. Displacement \(=8-4+4=8\) m
3. Distance \(=8+4+4=16\) m

Teacher's Note:
1. Rectangles below the axis count negative for displacement and positive for distance.
2. Do not mix up the order of the two answers.

Answer: (b) 45 m

Explanation:
1. 0 to 1 s: triangle, \(\tfrac12(1)(20)=10\) m
2. 1 to 2 s: rectangle, \(1\times20=20\) m
3. 2 to 3 s: trapezium from 20 to 10 m/s, \(\tfrac12(20+10)(1)=15\) m
4. Total \(=10+20+15=45\) m

Teacher's Note:
1. Use the area under the v-t graph for distance when velocity does not reverse.
2. Split into a triangle, rectangle and trapezium.

Answer: (d) 40 m

Explanation:
1. Acceleration is zero where the graph is horizontal, i.e. 0 to 20 s and 40 s to 60 s
2. Each interval has \(v=1\) m/s for 20 s, giving 20 m
3. Total displacement \(=20+20=40\) m

Teacher's Note:
1. Zero slope on a v-t graph means zero acceleration.
2. Ignore the middle triangular part, where the object is accelerating.

Answer: (c) -5 m/s²

Explanation:
1. At t = 2 s the velocity is 10 m/s and at t = 4 s it is 0
2. \(a_{av}=\dfrac{0-10}{4-2}=-5\) m/s²

Teacher's Note:
1. Average acceleration is change in velocity divided by the time interval.
2. The negative sign shows the block is slowing down.

Answer: (b) 30 m/s

Explanation:
1. 0 to 2 s: \(\Delta v=\tfrac12(2)(10)=10\) m/s
2. 2 s to 4 s: \(\Delta v=10\times2=20\) m/s
3. \(v(4)=0+10+20=30\) m/s

Teacher's Note:
1. Add the areas under the a-t graph to find the final velocity.
2. The particle started from rest, so the initial velocity is zero.