JEE Physics Motion In A Straight Line MCQs Set 13

Answer: (a) 4 : 9 and 2 : 3

Explanation:
1. \(H=\dfrac{u^2}{2g}\), so \(H_1:H_2=4:9\)
2. Time of flight \(=\dfrac{2u}{g}\), so \(T_1:T_2=2:3\)

Teacher's Note:
1. Height goes as the square of the speed, time of flight goes as the speed.
2. Both formulas follow from the equations of uniform acceleration.

Answer: (a) \(\sqrt2\,v_0\)

Explanation:
1. \(h=\dfrac{v_0^2}{2g}\), so \(v_0\propto\sqrt h\)
2. For 2h the velocity becomes \(\sqrt2\,v_0\)

Teacher's Note:
1. Height varies as the square of launch speed.
2. Doubling the height needs only a factor of \(\sqrt2\) in speed.

Answer: (d) 68.6 m/s

Explanation:
1. Equal distances in consecutive seconds on either side of the highest point means the body is at the top at the end of the 7th second
2. So \(u=gt=9.8\times7\)
3. \(u=68.6\) m/s

Teacher's Note:
1. The motion is symmetric about the highest point.
2. The 7th second is the second ending at t = 7 s.

Answer: (b) \(\tfrac12gt^2\)

Explanation:
1. Reverse the motion: the last t seconds of ascent is the first t seconds of a fall from rest at the top
2. Distance \(=\tfrac12gt^2\)

Teacher's Note:
1. Time reversal turns the last part of the rise into a free fall from rest.
2. The answer does not depend on u.

Answer: (b) 11 : 1

Explanation:
1. \(u=6g\)
2. First second: \(s_1=u-\tfrac g2=5.5g\)
3. The seventh second is the first second of fall from the top: \(s_7=\tfrac g2=0.5g\)
4. Ratio \(=11:1\)

Teacher's Note:
1. The 7th second starts at the top, so it is a free fall from rest.
2. Use the distance in the nth second, \(u-\tfrac g2(2n-1)\), only while the body is rising.

Answer: (a) 20 m

Explanation:
1. By time reversal, the last 2 s of the rise equals a free fall of 2 s from rest
2. \(s=\tfrac12(10)(2)^2=20\) m

Teacher's Note:
1. The initial speed 600 m/s is irrelevant for the last part of the rise.
2. The result matches the first 2 s of any free fall.

Answer: (d) 12 s

Explanation:
1. Equal heights: \(98t-4.9t^2=98(t-4)-4.9(t-4)^2\)
2. This gives \(392=4.9(8t-16)\), so \(8t-16=80\)
3. \(t=12\) s

Teacher's Note:
1. Both bodies are at the same height when they meet.
2. Measure the second body's time from its own launch, which is t - 4.

Answer: (b) the time of ascent is less than the time of descent

Explanation:
1. While rising, gravity and air resistance both act downwards, so the retardation is \(g+f/m\)
2. While falling, the acceleration is \(g-f/m\), which is smaller
3. The same distance at a smaller acceleration takes longer, so ascent time is less than descent time

Teacher's Note:
1. Air resistance always opposes the motion.
2. So the body decelerates faster on the way up than it accelerates on the way down.

Answer: (c) \(v_1>v_2\)

Explanation:
1. Air resistance removes mechanical energy on both the upward and the downward trip
2. So the kinetic energy on return is less than the initial kinetic energy
3. Hence \(v_2\lt v_1\)

Teacher's Note:
1. Air drag is a non-conservative force, so energy is lost.
2. With no drag the speeds would be equal.

Answer: (b) 0, 10 m/s

Explanation:
1. After one full lap the displacement is zero, so average velocity = 0
2. Average speed \(=\dfrac{2\pi\times100}{62.8}=10\) m/s

Teacher's Note:
1. Over a complete lap the displacement is zero but the distance is the circumference.
2. Average speed is therefore the same as the constant speed of the car.

Answer: (d) 16 m

Explanation:
1. \(v=12-3t^2=0\) gives \(t=2\) s
2. \(x(0)=40\) m and \(x(2)=40+24-8=56\) m
3. Distance travelled before coming to rest \(=56-40=16\) m

Teacher's Note:
1. The distance travelled is measured from the starting position x = 40 m, not from the origin.
2. The motion does not reverse before t = 2 s, so distance equals displacement here.

Answer: (d) \(\dfrac45\)

Explanation:
1. \(t=\sqrt{2h/g}\), independent of mass
2. \(t_A:t_B=\sqrt{16}:\sqrt{25}=4:5\)

Teacher's Note:
1. Free fall time does not depend on mass.
2. Time varies as the square root of the height.

Answer: (b) \(\dfrac{2v_dv_u}{v_d+v_u}\)

Explanation:
1. Let XY = d: time \(=\dfrac d{v_u}+\dfrac d{v_d}\)
2. Average speed \(=\dfrac{2d}{d\left(\frac1{v_u}+\frac1{v_d}\right)}=\dfrac{2v_uv_d}{v_u+v_d}\)

Teacher's Note:
1. Total distance over total time gives the harmonic mean for equal distances.
2. Do not take the simple mean of the two speeds.

Answer: (a) \(\tfrac12f_0T\)

Explanation:
1. \(f=0\) at \(t=T\)
2. \(v=\int_0^T f_0\left(1-\dfrac tT\right)dt=f_0\left[T-\dfrac T2\right]\)
3. \(v=\tfrac12f_0T\)

Teacher's Note:
1. Find the time at which acceleration vanishes, then integrate up to that time.
2. The units of \(f_0T\) are those of velocity, which supports the answer.

Answer: (b) 9

Explanation:
1. Average velocity \(=\dfrac{10+20}{2}=15\) m/s
2. \(t=\dfrac{135}{15}=9\) s

Teacher's Note:
1. With constant acceleration, average velocity is the mean of initial and final velocity.
2. This avoids solving for the acceleration first.

Answer: (c) \(3S_1\)

Explanation:
1. From rest, distances in equal successive intervals are in the ratio 1 : 3
2. So \(S_2=3S_1\)

Teacher's Note:
1. Equal time intervals from rest give distances in the ratio 1 : 3 : 5 : ...
2. Check: \(S_1=50a\), \(S_2=200a-50a=150a\).

Answer: (b) 20 m/s

Explanation:
1. Relative speed needed \(=\dfrac{1000}{100}=10\) m/s
2. Scooter speed \(=10+10=20\) m/s

Teacher's Note:
1. Overtaking depends on relative speed.
2. The scooterist must close 1000 m of gap at 10 m/s relative speed.

Answer: (b) 75 m/s

Explanation:
1. First ball falls for 18 s: \(\tfrac12(10)(18)^2=1620\) m
2. Second ball moves for 12 s: \(12v+\tfrac12(10)(12)^2=12v+720\)
3. Equate: \(12v+720=1620\), so \(v=75\) m/s

Teacher's Note:
1. At the meeting instant both have covered the same distance from the platform.
2. The second ball has been moving for 18 - 6 = 12 s.

Answer: (c) \(7\sqrt2\) units

Explanation:
1. \(\vec v=\vec u+\vec at=(3+4)\hat i+(4+3)\hat j=7\hat i+7\hat j\)
2. Speed \(=\sqrt{49+49}=7\sqrt2\)

Teacher's Note:
1. Apply \(\vec v=\vec u+\vec at\) component by component.
2. Speed is the magnitude of the final velocity vector.

Answer: (d) 5 m/s²

Explanation:
1. \(\Delta\vec v=40\hat j-30\hat i\), so \(|\Delta\vec v|=\sqrt{40^2+30^2}=50\) m/s
2. Average acceleration \(=\dfrac{50}{10}=5\) m/s²

Teacher's Note:
1. The change of velocity is a vector difference.
2. The simple difference 40 - 30 = 10 would be wrong.

Answer: (b) \(\dfrac2v=\dfrac1{v_1}+\dfrac1{v_2}\)

Explanation:
1. Total displacement \(=2x\), total time \(=\dfrac x{v_1}+\dfrac x{v_2}\)
2. \(v=\dfrac{2x}{x\left(\frac1{v_1}+\frac1{v_2}\right)}\), so \(\dfrac2v=\dfrac1{v_1}+\dfrac1{v_2}\)

Teacher's Note:
1. Equal distances give a relation between reciprocals of velocities.
2. The reciprocal of the average velocity is the mean of reciprocals.