JEE Physics Motion In A Straight Line MCQs Set 16

Answer: (c) 9.8 m

Explanation:
1. Both have the same acceleration g and the same initial velocity (zero)
2. Relative acceleration is zero and relative velocity is zero, so the separation stays 9.8 m

Teacher's Note:
1. Bodies in free fall have zero relative acceleration.
2. The separation changes only if the initial velocities differ.

Answer: (c) 20 m above the ground

Explanation:
1. Relative acceleration is zero; the balls approach at the constant relative speed \(20+20=40\) m/s
2. Time to meet \(t=\dfrac{80}{40}=2\) s
3. Height of A \(=20(2)-\tfrac12(10)(4)=20\) m above the ground

Teacher's Note:
1. In free fall both balls have the same acceleration, so the relative motion is uniform.
2. Find the meeting time first, then the position.

Answer: (b) 10 m/s

Explanation:
1. First walnut: total fall time \(t=\sqrt{\dfrac{2(20)}{10}}=2\) s, and after 5 m, \(t=1\) s have passed
2. The second walnut is thrown at that instant and has 15 m to fall in the remaining 1 s
3. \(15=u(1)+\tfrac12(10)(1)^2\Rightarrow u=10\) m/s

Teacher's Note:
1. Both walnuts must hit the ground at 2 s after the first drop.
2. The second one gets only the remaining 1 s.

Answer: (b) \(\sqrt{gH}\)

Explanation:
1. Relative acceleration is zero, so the bodies close the gap H at the constant speed \(v_0\): meeting time \(t=\dfrac H{v_0}\)
2. Body B falls \(\tfrac12gt^2=\dfrac H2\) when they meet at height \(\dfrac H2\), so \(t^2=\dfrac Hg\)
3. \(\dfrac{H^2}{v_0^2}=\dfrac Hg\Rightarrow v_0=\sqrt{gH}\)

Teacher's Note:
1. Dropping and throwing simultaneously gives uniform relative motion.
2. Equate the fall of B to H/2 to find the time.

Answer: (b) 1

Explanation:
1. Relative to the elevator the bolt accelerates downwards with \(32+6=38\) ft/s² from zero relative speed
2. \(19=\tfrac12(38)t^2\Rightarrow t^2=1\)
3. \(t=1\) s

Teacher's Note:
1. In an accelerating lift, the effective downward acceleration is \(g+a\).
2. The bolt starts with zero velocity relative to the lift.

Answer: (a) 24 m/s

Explanation:
1. Total distance \(=21\times3600=75600\) m
2. First part \(=14\times1080=15120\) m
3. Remaining distance \(=60480\) m in \(2520\) s, so speed \(=\dfrac{60480}{2520}=24\) m/s

Teacher's Note:
1. Average speed is total distance over total time.
2. Convert minutes to seconds before multiplying.

Answer: (b) 8

Explanation:
1. \(v=8-8t\) is zero at t = 1 s, so the particle turns back there
2. \(x(0)=2\), \(x(1)=6\), \(x(2)=2\)
3. Distance \(=4+4=8\) m

Teacher's Note:
1. Find the turning point before adding distances.
2. The displacement here is zero, but the distance is not.

Answer: (c) \(v^3\)

Explanation:
1. \(s=k\sqrt t\Rightarrow v=\dfrac{ds}{dt}=\dfrac k2t^{-1/2}\)
2. \(a=\dfrac{dv}{dt}=-\dfrac k4t^{-3/2}\)
3. Since \(v^3\propto t^{-3/2}\), \(a\propto v^3\)

Teacher's Note:
1. Differentiate step by step and compare the powers of t.
2. The negative sign only shows that the particle is retarding.

Answer: (d) \(\left(\dfrac{2v_0^2}{\alpha}\right)^{1/4}\)

Explanation:
1. \(v\dfrac{dv}{dx}=-\alpha x^3\), so \(\int_{v_0}^{0}v\,dv=-\alpha\int_0^{x}x^3dx\)
2. \(-\dfrac{v_0^2}{2}=-\dfrac{\alpha x^4}{4}\)
3. \(x=\left(\dfrac{2v_0^2}{\alpha}\right)^{1/4}\)

Teacher's Note:
1. When acceleration depends on position, use \(a=v\,dv/dx\).
2. Check the units: \(\alpha x^4\) must have the units of \(v^2\).

Answer: (a) 8.3 m

Explanation:
1. \(s=ut+\tfrac12at^2\): \(5u+12.5a=10\) and \(8u+32a=20\)
2. Solving: \(a=\dfrac13\) m/s² and \(u=\dfrac76\) m/s
3. Distance in the next 2 s \(=s(10)-s(8)=\left(\dfrac{70}{6}+\dfrac{50}{3}\right)-20=\dfrac{25}{3}\approx8.3\) m

Teacher's Note:
1. Use the distance equation at the three times 5 s, 8 s and 10 s.
2. Solve for u and a first.

Answer: (c) 3/5

Explanation:
1. Let the maximum speed be v. Time for the first stage \(=\dfrac{S}{v/2}=\dfrac{2S}v\); second \(=\dfrac{2S}v\); third \(=\dfrac{3S}{v/2}=\dfrac{6S}v\)
2. Total time \(=\dfrac{10S}v\) and total distance \(=6S\)
3. Average velocity \(=\dfrac{6S}{10S/v}=\dfrac35v\)

Teacher's Note:
1. In uniform acceleration from rest, the average speed is half the final speed.
2. Add the times of all three stages for the total.

Answer: (a) 3 : 2

Explanation:
1. \(v_M^2=\dfrac{v_A^2+v_B^2}{2}=\dfrac{49+289}{2}=169\), so \(v_M=13\) m/s
2. \(t_1=\dfrac{13-7}{a}=\dfrac6a\) and \(t_2=\dfrac{17-13}{a}=\dfrac4a\)
3. \(t_1:t_2=3:2\)

Teacher's Note:
1. The speed at the mid-point of the distance is the root mean square of the end speeds.
2. Equal changes of velocity take equal times, which is not the case here.

Answer: (a) More than 19.6 m/s

Explanation:
1. Time of flight of one ball \(=\dfrac{2u}{g}\)
2. For more than two balls to be in the air, the first ball must still be flying when the third is thrown, i.e. after 4 s: \(\dfrac{2u}{g}>4\)
3. \(u>2g=19.6\) m/s

Teacher's Note:
1. Count the balls in the air at the moment the third ball is thrown.
2. Time of flight must exceed twice the throwing interval.

Answer: (b) \(2\left(\dfrac ug-t_1\right)\)

Explanation:
1. The time to reach the top is \(\dfrac ug\), so the body is at the top after \(\dfrac ug-t_1\) more seconds
2. By symmetry it takes the same time to come back to the point
3. Time between the two passages \(=2\left(\dfrac ug-t_1\right)\)

Teacher's Note:
1. Rising and falling parts between the same two heights take equal times.
2. Measure the time from the first passage of the point.

Answer: (c) \(t=\sqrt{t_1t_2}\)

Explanation:
1. Take height h and throw speed u; the displacement to the ground is \(-h\)
2. Upward throw: \(-h=ut_1-\tfrac12gt_1^2\); downward throw: \(-h=-ut_2-\tfrac12gt_2^2\). Subtracting gives \(u=\tfrac12g(t_1-t_2)\) and then \(h=\tfrac12gt_1t_2\)
3. Free fall: \(h=\tfrac12gt^2\Rightarrow t=\sqrt{t_1t_2}\)

Teacher's Note:
1. Write the displacement equation for both throws and eliminate u.
2. The final relation is symmetric in the two times.

Answer: (c) reach the ground in 4 s

Explanation:
1. At release the balloon is at \(\tfrac12(1.25)(64)=40\) m and moving up with \(v=1.25\times8=10\) m/s
2. The stone starts with 10 m/s upwards: \(-40=10t-5t^2\Rightarrow t^2-2t-8=0\Rightarrow t=4\) s
3. The displacement is 40 m downwards and the distance is 50 m (5 m up, 45 m down), and it first moves up, so only (c) is correct

Teacher's Note:
1. A released object keeps the velocity of the carrier at that moment.
2. Check each statement with the numbers found.

Answer: (a) 5 m/s

Explanation:
1. Free fall for 10 s: \(v=100\) m/s and \(h=500\) m
2. Remaining height \(=2495-500=1995\) m
3. \(v^2=100^2-2(2.5)(1995)=10000-9975=25\Rightarrow v=5\) m/s

Teacher's Note:
1. Two stages: free fall then retarded motion.
2. The end speed of stage one is the start speed of stage two.

Answer: (b) 10 s

Explanation:
1. Positions from Manav's start: \(x_M=28t\) and \(x_Y=80+2t^2\)
2. Equal when \(2t^2-28t+80=0\Rightarrow t^2-14t+40=0\Rightarrow t=4\) s or 10 s
3. At 4 s Manav catches up with Yogesh (Manav overtakes); then Yogesh speeds up and overtakes Manav at 10 s

Teacher's Note:
1. The two roots mean they meet twice.
2. Decide who is ahead before and after each meeting.

Answer: (c) 16 s

Explanation:
1. Speed of A \(=\tan37^\circ=\dfrac34\) m/s; speed of B \(=\tan53^\circ=\dfrac43\) m/s
2. \(x_A=4+\dfrac34t\) and \(x_B=\dfrac43(t-4)\)
3. Equal when \(4+\dfrac34t=\dfrac43t-\dfrac{16}3\Rightarrow\dfrac{7}{12}t=\dfrac{28}{3}\Rightarrow t=16\) s after A starts

Teacher's Note:
1. The slope of a position-time graph is the velocity.
2. The question asks for the time measured from the start of A, not of B.

Answer: (a) \(\sqrt{\dfrac h{8g}}\)

Explanation:
1. For the stone reaching 4h: \(u^2=2g(4h)\Rightarrow u=\sqrt{8gh}\)
2. Relative acceleration is zero and the relative speed is u, so they meet when \(t=\dfrac hu\)
3. \(t=\dfrac h{\sqrt{8gh}}=\sqrt{\dfrac h{8g}}\)

Teacher's Note:
1. Two bodies in free fall approach each other at constant speed.
2. Find u from the maximum height first.

Answer: (c) 31.25 m

Explanation:
1. Let the first stone take T seconds: \(h=5T^2\)
2. The second stone falls \(h-20=5(T-1)^2\) in \(T-1\) s
3. \(5T^2-20=5T^2-10T+5\Rightarrow T=2.5\) s, so \(h=5(2.5)^2=31.25\) m

Teacher's Note:
1. Subtract the two equations to eliminate \(T^2\).
2. The second stone has one second less to fall and 20 m less height.

Answer: (d) C

Explanation:
1. Velocity is the slope of the distance-time curve
2. The curve is flat near A, becomes steeper through B, is steepest at C (the inflection point) and flattens again towards D
3. The maximum slope is near C

Teacher's Note:
1. The steepest part of the s-t graph gives the maximum speed.
2. At the inflection point the slope stops increasing.

Answer: (c) 54 m

Explanation:
1. \(v=18t-3t^2\) and \(a=18-6t\)
2. Maximum speed when \(a=0\Rightarrow t=3\) s
3. \(x=9(9)-27=54\) m

Teacher's Note:
1. Maximum or minimum velocity occurs where the acceleration is zero.
2. Substitute the time back into the position equation.

Answer: (d) 16 m

Explanation:
1. \(v=12-3t^2=0\Rightarrow t=2\) s
2. \(x(0)=40\) and \(x(2)=40+24-8=56\)
3. Distance \(=56-40=16\) m

Teacher's Note:
1. The particle comes to rest when its velocity is zero.
2. Distance is the difference of positions, not the position itself.