JEE Physics Motion In A Plane MCQs Set 11

Answer: (b) vertical velocity versus t:

Explanation:
1. Acceleration is constant (-g), so graph (a) is correct
2. Horizontal displacement grows linearly with time, so (c) is correct
3. Vertical displacement is a downward-opening parabola, so (d) is correct
4. Vertical velocity \(v_y=u_y-gt\) is a straight line that crosses zero and becomes negative, not a parabola touching the axis, so (b) is incorrect

Teacher's Note:
1. Vertical velocity changes linearly with time because the acceleration is constant.
2. A parabola in the v-t graph would mean a changing acceleration.

Answer: (c) is perpendicular to the acceleration for one instant only

Explanation:
1. The acceleration is g, vertically downwards
2. Velocity is perpendicular to it only when its vertical component is zero, which happens once, at the highest point

Teacher's Note:
1. At the top of the path the velocity is purely horizontal.
2. The angle between velocity and acceleration is 90 degrees only there.

Answer: (a) constant velocity

Explanation:
1. With no air resistance there is no horizontal force
2. Hence the horizontal acceleration is zero and the horizontal velocity is constant

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. Only gravity acts, and it acts vertically.

Answer: (d) 50 s

Explanation:
1. Vertical component \(u_y=1000\sin30^\circ=500\) m/s
2. \(t=\dfrac{u_y}{g}=\dfrac{500}{10}=50\) s

Teacher's Note:
1. The time to the highest point is half the time of flight.
2. Only the vertical component matters.

Answer: (a) 125 m

Explanation:
1. \(T=\dfrac{2u_y}g\Rightarrow u_y=\dfrac{10\times10}{2}=50\) m/s
2. \(H=\dfrac{u_y^2}{2g}=\dfrac{2500}{20}=125\) m

Teacher's Note:
1. The range is not needed to find the height once the time of flight is known.
2. Use the vertical motion only.

Answer: (b) 2 s

Explanation:
1. \(T=\dfrac{2u\sin\theta}g=\dfrac{2(19.6)(0.5)}{9.8}=2\) s

Teacher's Note:
1. Time of flight depends on the vertical component of the initial velocity.
2. The formula applies for landing at the launch level.

Answer: (d) \(\dfrac{u^2\cos^2\theta}{2g}\)

Explanation:
1. The angle with the horizontal is \(90^\circ-\theta\), so the vertical component is \(u\sin(90^\circ-\theta)=u\cos\theta\)
2. \(H=\dfrac{(u\cos\theta)^2}{2g}=\dfrac{u^2\cos^2\theta}{2g}\)

Teacher's Note:
1. Read carefully whether the angle is measured from the horizontal or the vertical.
2. The sine and cosine swap when the reference line changes.

Answer: (b) 21.1 m

Explanation:
1. Both players release and catch the ball at 1.5 m, so the flight is symmetric with total time 4 s and rise time 2 s
2. Rise above the release point \(=\tfrac12g(2)^2=19.6\) m
3. Maximum height from the ground \(=19.6+1.5=21.1\) m

Teacher's Note:
1. Do not forget to add the height of the throwing point.
2. The rise time is half of the total flight time.

Answer: (c) \(\sqrt{v^2+g^2t^2-(2v\sin\theta)gt}\)

Explanation:
1. \(v_x=v\cos\theta\) and \(v_y=v\sin\theta-gt\)
2. Speed\(^2=v^2\cos^2\theta+v^2\sin^2\theta-2vgt\sin\theta+g^2t^2\)
3. Speed \(=\sqrt{v^2+g^2t^2-(2v\sin\theta)gt}\)

Teacher's Note:
1. Only the vertical component changes with time.
2. Expand the square of \(v_y\) carefully.

Answer: (c) \(2u\sin\theta\)

Explanation:
1. Initial velocity \((u\cos\theta,\,u\sin\theta)\) and final velocity \((u\cos\theta,\,-u\sin\theta)\)
2. Change \(=(0,\,-2u\sin\theta)\), magnitude \(2u\sin\theta\)

Teacher's Note:
1. The horizontal component does not change.
2. The vertical component reverses its direction.

Answer: (c) 30°

Explanation:
1. \(u_x=6\) m/s. Vertical: \(v_y^2=u_y^2-2gh\Rightarrow4=u_y^2-2(10)(0.4)\Rightarrow u_y^2=12\)
2. \(u_y=2\sqrt3\) m/s
3. \(\tan\theta=\dfrac{2\sqrt3}{6}=\dfrac1{\sqrt3}\Rightarrow\theta=30^\circ\)

Teacher's Note:
1. Use the vertical motion equation to get the launch velocity from the velocity at a height.
2. The horizontal component is constant.

Answer: (b) \(\dfrac{\sqrt3u^2}{2g}\)

Explanation:
1. \(R=\dfrac{u^2\sin2\theta}g=\dfrac{u^2\sin120^\circ}g\)
2. \(\sin120^\circ=\dfrac{\sqrt3}{2}\), so \(R=\dfrac{\sqrt3u^2}{2g}\)

Teacher's Note:
1. The range formula is \(R=u^2\sin2\theta/g\).
2. Option (d) has the wrong dimensions.

Answer: (b) 9.8 m

Explanation:
1. \(T=\dfrac{2u_y}g=\dfrac{2(4.9)}{9.8}=1\) s
2. \(R=u_xT=9.8\times1=9.8\) m

Teacher's Note:
1. Range equals \(u_x\times T\).
2. The time of flight comes from the vertical component alone.

Answer: (d) 12 m

Explanation:
1. \(u_x=5\) m/s and \(u_y=12\) m/s
2. \(T=\dfrac{2(12)}{10}=2.4\) s
3. \(R=5\times2.4=12\) m

Teacher's Note:
1. Read the components directly from the vector.
2. Use the time of flight from the vertical component.

Answer: (c) \(\dfrac{2v_1v_2}{g}\)

Explanation:
1. For the person on shore the shell has horizontal velocity \(v_2\) (the ship's speed) and vertical velocity \(v_1\)
2. \(T=\dfrac{2v_1}g\)
3. \(R=v_2T=\dfrac{2v_1v_2}g\)

Teacher's Note:
1. The shell shares the ship's horizontal velocity.
2. Check the dimensions: only (c) has the dimension of length.

Answer: (b) \(12.5\times10^3\) m

Explanation:
1. \(R=\dfrac{u^2\sin2\theta}g=\dfrac{(500)^2\sin30^\circ}{10}\)
2. \(=\dfrac{250000\times0.5}{10}=12500\) m \(=12.5\times10^3\) m

Teacher's Note:
1. \(\sin30^\circ=\tfrac12\) makes the calculation easy.
2. Check the powers of ten in the options carefully.

Answer: (a) 9 times

Explanation:
1. \(R=\dfrac{u^2\sin2\theta}g\propto u^2\)
2. Tripling u multiplies the range by \(3^2=9\)

Teacher's Note:
1. Range varies as the square of the speed.
2. Doubling the speed would give four times the range.

Answer: (a) 1

Explanation:
1. Range is maximum when the angle with the horizontal is 45°
2. Then the angle with the vertical is also 45°, so \(\tan\theta=1\)

Teacher's Note:
1. Maximum range occurs at 45° to the horizontal.
2. This is also 45° to the vertical.

Answer: (b) 500 m/s

Explanation:
1. Maximum range \(R_{max}=\dfrac{u^2}g\)
2. \(u^2=25000\times10=250000\)
3. \(u=500\) m/s

Teacher's Note:
1. Convert kilometres to metres first.
2. Maximum range corresponds to 45° projection.

Answer: (c) \(20\sqrt2\) m

Explanation:
1. The maximum jump is at 45°: \(\dfrac{u^2}g=1.6\Rightarrow u=4\) m/s
2. Time of one jump \(T=\dfrac{2u\sin45^\circ}g=\dfrac{\sqrt2\times4}{10}\) s \(=0.4\sqrt2\) s
3. Number of jumps in 10 s \(=\dfrac{10}{0.4\sqrt2}=\dfrac{25}{\sqrt2}\), so distance \(=\dfrac{25}{\sqrt2}\times1.6=20\sqrt2\) m

Teacher's Note:
1. Each jump is a projectile at 45° for the maximum distance.
2. Divide the total time by the time of one jump.

Answer: (b) The launch speed is the same for all the three

Explanation:
1. Same maximum height means the same vertical launch velocity, hence the same time of flight, so (a) and (d) are correct
2. The ranges are different in the same time, so the horizontal velocity is largest for C (c is correct)
3. The launch speeds \(\sqrt{u_x^2+u_y^2}\) differ because \(u_x\) differs, so (b) is incorrect

Teacher's Note:
1. Height and time of flight depend only on the vertical component.
2. Range then depends on the horizontal component.

Answer: (b) \(b=a\)

Explanation:
1. \(R=\dfrac{2ab}g\) and \(H=\dfrac{b^2}{2g}\)
2. \(R=4H\Rightarrow\dfrac{2ab}g=\dfrac{2b^2}g\)
3. \(a=b\)

Teacher's Note:
1. Read the horizontal and vertical components from the vector.
2. R = 4H corresponds to a launch angle of 45°.

Answer: (a) Their horizontal ranges will be the same

Explanation:
1. Angles 60° and 30° are complementary, so \(\sin120^\circ=\sin60^\circ\) and the ranges are equal
2. Heights \(\propto\sin^2\theta\) and times \(\propto\sin\theta\) differ
3. Landing velocities have different directions

Teacher's Note:
1. Complementary angles give equal ranges for the same speed.
2. Heights and times of flight differ for complementary angles.