JEE Physics Motion In A Plane MCQs Set 12

Answer: (d) \(\pi\left(\dfrac{u^2}g\right)^2\)

Explanation:
1. The farthest bullets land at the maximum range \(R_{max}=\dfrac{u^2}g\)
2. The ground covered is a circle of this radius, \(A=\pi R_{max}^2=\pi\left(\dfrac{u^2}g\right)^2\)

Teacher's Note:
1. The region reached on the ground is a circle around the gun.
2. The radius is the maximum range.

Answer: (a) P

Explanation:
1. \(R\propto\sin2\theta\): \(\sin30^\circ=0.5\), \(\sin60^\circ=0.87\), \(\sin90^\circ=1\), \(\sin120^\circ=0.87\)
2. The smallest value is for 15°, so P has the shortest range

Teacher's Note:
1. Compare the values of \(\sin2\theta\).
2. The angle farthest from 45° gives the shortest range.

Answer: (d) 20 m

Explanation:
1. Maximum height (vertical throw) \(=\dfrac{u^2}{2g}=10\Rightarrow u^2=20g\)
2. Maximum range \(=\dfrac{u^2}g=20\) m

Teacher's Note:
1. The maximum range is twice the maximum height reached in a vertical throw.
2. The boy throws with the same speed in both cases.

Answer: (d) 50 m

Explanation:
1. \(R_{max}=\dfrac{u^2}g=100\) m
2. Maximum height (vertical throw) \(=\dfrac{u^2}{2g}=50\) m

Teacher's Note:
1. The maximum height is half the maximum range for the same speed.
2. Throwing vertically gives the greatest height.

Answer: (b) 500 m

Explanation:
1. For the maximum range \(\theta=45^\circ\) and \(R_{max}=\dfrac{u^2}g=2000\) m
2. \(H=\dfrac{u^2\sin^2 45^\circ}{2g}=\dfrac{u^2}{4g}=\dfrac{2000}4=500\) m

Teacher's Note:
1. At 45° the maximum height is one quarter of the range.
2. Do not confuse this with the height of a vertical throw.

Answer: (c) beyond the enemy target

Explanation:
1. \(R\propto\sin2\theta\) for the same speed
2. \(\sin60^\circ=0.866>\sin40^\circ=0.643\), and both angles are below 45°
3. The range increases, so the shell falls beyond the target

Teacher's Note:
1. Below 45°, a larger angle gives a larger range.
2. The range is largest at 45°.

Answer: (a) \(\dfrac R3\)

Explanation:
1. \(\dfrac HR=\dfrac{\tan\theta}4=\dfrac{4/3}{4}=\dfrac13\)
2. \(H=\dfrac R3\)

Teacher's Note:
1. The relation \(H=\dfrac{R\tan\theta}{4}\) is very useful.
2. \(\tan53^\circ=\dfrac43\).

Answer: (a) \(h_1h_2\propto R^2\)

Explanation:
1. The two angles are \(\theta\) and \(90^\circ-\theta\)
2. \(h_1h_2=\dfrac{u^2\sin^2\theta}{2g}\cdot\dfrac{u^2\cos^2\theta}{2g}=\dfrac{u^4\sin^2\theta\cos^2\theta}{4g^2}\)
3. Since \(R=\dfrac{2u^2\sin\theta\cos\theta}g\), \(h_1h_2=\dfrac{R^2}{16}\propto R^2\)

Teacher's Note:
1. Same speed and same range need complementary angles.
2. Express \(\sin\theta\cos\theta\) through R.

Answer: (d) 160 m

Explanation:
1. \(H\propto\sin^2\theta\)
2. \(\dfrac{H_{53}}{H_{37}}=\dfrac{(0.8)^2}{(0.6)^2}=\dfrac{16}9\)
3. \(H_{53}=90\times\dfrac{16}9=160\) m

Teacher's Note:
1. \(\sin37^\circ=0.6\) and \(\sin53^\circ=0.8\).
2. Maximum height is proportional to \(\sin^2\theta\) for fixed speed.

Answer: (a) \(3y_1\)

Explanation:
1. Equal ranges mean complementary angles, so the other angle is \(\dfrac\pi3\)
2. \(\dfrac{y_2}{y_1}=\dfrac{\sin^2 60^\circ}{\sin^2 30^\circ}=\dfrac{3/4}{1/4}=3\)
3. \(y_2=3y_1\)

Teacher's Note:
1. Complementary angles have equal ranges.
2. Compare the heights using \(\sin^2\theta\).

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

Explanation:
1. The angles are \(\theta\) and \(90^\circ-\theta\)
2. \(y_1+y_2=\dfrac{u^2\sin^2\theta}{2g}+\dfrac{u^2\cos^2\theta}{2g}=\dfrac{u^2}{2g}\)

Teacher's Note:
1. \(\sin^2\theta+\cos^2\theta=1\).
2. The sum equals the height of a vertical throw.

Answer: (c) 45°

Explanation:
1. \(\dfrac RH=\dfrac4{\tan\theta}=4\)
2. \(\tan\theta=1\Rightarrow\theta=45^\circ\)

Teacher's Note:
1. \(R=4H\) is the signature of 45° projection.
2. Use \(R/H=4\cot\theta\).

Answer: (d) 30°

Explanation:
1. \(\dfrac RH=\dfrac4{\tan\theta}=4\sqrt3\)
2. \(\tan\theta=\dfrac1{\sqrt3}\Rightarrow\theta=30^\circ\)

Teacher's Note:
1. Use \(R/H=4\cot\theta\).
2. \(\tan30^\circ=1/\sqrt3\).

Answer: (a) \(\tan^{-1}2\)

Explanation:
1. \(\dfrac RH=4\cot\theta\)
2. \(R=2H\Rightarrow4\cot\theta=2\Rightarrow\tan\theta=2\)
3. \(\theta=\tan^{-1}2\)

Teacher's Note:
1. A range equal to H would correspond to \(\tan^{-1}4\).
2. Use \(R/H=4\cot\theta\).

Answer: (c) \(4\sqrt3\)

Explanation:
1. \(u\cos\theta=\dfrac{\sqrt3}2u\Rightarrow\theta=30^\circ\)
2. \(\dfrac RH=4\cot30^\circ=4\sqrt3\)

Teacher's Note:
1. Speed at the top is \(u\cos\theta\).
2. \(\cot30^\circ=\sqrt3\).

Answer: (d) 45°

Explanation:
1. Moving horizontally at the top of the wall means the wall top is the highest point: \(H=45\) m and half-range \(=90\) m, so \(R=180\) m
2. \(\tan\theta=\dfrac{4H}R=\dfrac{180}{180}=1\)
3. \(\theta=45^\circ\)

Teacher's Note:
1. Horizontal velocity at the wall means the shell is at its highest point there.
2. Use \(\tan\theta=4H/R\).

Answer: (c) (100, 50) m

Explanation:
1. Maximum range means \(\theta=45^\circ\) and \(R=200\) m
2. The speed is minimum at the highest point, which is at \(x=\dfrac R2=100\) m
3. \(H=\dfrac R4=50\) m, so the point is (100, 50) m

Teacher's Note:
1. The minimum speed occurs at the top of the trajectory.
2. At the top x is half the range.

Answer: (b) \(\dfrac12\sin^{-1}\left(\dfrac{\sqrt3}8\right)\)

Explanation:
1. Equal ranges: \(\dfrac{u^2\sin2\theta_A}g=\dfrac{(u/2)^2\sin60^\circ}g\)
2. \(\sin2\theta_A=\dfrac14\cdot\dfrac{\sqrt3}2=\dfrac{\sqrt3}8\)
3. \(\theta_A=\dfrac12\sin^{-1}\left(\dfrac{\sqrt3}8\right)\)

Teacher's Note:
1. Equate the two range expressions.
2. The result must be a small angle since A is much faster.

Answer: (c) \(\sqrt3:2\)

Explanation:
1. Equal times of flight: \(\dfrac{2v_1}g=\dfrac{2v_2\sin60^\circ}g\)
2. \(\dfrac{v_1}{v_2}=\sin60^\circ=\dfrac{\sqrt3}2\)

Teacher's Note:
1. Equal flight times mean equal vertical launch velocities.
2. The first ball's whole speed is vertical.

Answer: (b) \(\tan^2\theta:1\)

Explanation:
1. First body: \(H_1=\dfrac{u^2\sin^2\theta}{2g}\)
2. Second body makes \(90^\circ-\theta\) with the horizontal: \(H_2=\dfrac{u^2\cos^2\theta}{2g}\)
3. \(\dfrac{H_1}{H_2}=\tan^2\theta:1\)

Teacher's Note:
1. An angle with the vertical swaps sine and cosine.
2. Maximum height varies as the square of the vertical component.

Answer: (a) 5 : 4

Explanation:
1. \(H=\dfrac{u_y^2}{2g}\) and \(T=\dfrac{2u_y}g\)
2. \(\dfrac H{T^2}=\dfrac{u_y^2/2g}{4u_y^2/g^2}=\dfrac g8=\dfrac{10}8=\dfrac54\)

Teacher's Note:
1. Both H and T depend on the vertical component, which cancels out.
2. The ratio is \(g/8\) for every projectile.

Answer: (a) \(\dfrac E4\)

Explanation:
1. At the top only the horizontal velocity \(u\cos60^\circ=\dfrac u2\) remains
2. \(KE=\dfrac12m\left(\dfrac u2\right)^2=\dfrac E4\)

Teacher's Note:
1. Kinetic energy depends on the square of the speed.
2. The vertical part of the kinetic energy turns into potential energy.

Answer: (b) K/4

Explanation:
1. 30° with the vertical means 60° with the horizontal
2. At the top \(KE=K\cos^2 60^\circ=\dfrac K4\)

Teacher's Note:
1. Convert the angle to the horizontal first.
2. At the top the speed equals the horizontal component.