JEE Physics Projectile Motion MCQs Set 01

Answer: (c) \( 30^\circ \)

Explanation:
1. Height \( H = u^2\sin^2\theta/2g \) is equal, so \( u_A^2\sin^245^\circ = 2u_A^2\sin^2\theta_B \).
2. \( \sin^2\theta_B = 1/4 \), so \( \theta_B = 30^\circ \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (b) \( 22.36\text{ m s}^{-1},\ \tan^{-1}2 \)

Explanation:
1. At t = 2 s the vertical velocity is zero, so \( u_y = 20 \) m/s. At t = 1 s, \( v_y = 10 = v_x \) (45°).
2. \( u = \sqrt{10^2+20^2} = 22.36 \) m/s and \( \tan\theta = 20/10 = 2 \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( \dfrac{3mv^3}{16g} \)

Explanation:
1. At the top the velocity is \( v\cos60^\circ = v/2 \) horizontally and the height is \( H = v^2\sin^260^\circ/2g = 3v^2/8g \).
2. \( L = mv_xH = m\cdot\dfrac v2\cdot\dfrac{3v^2}{8g} = \dfrac{3mv^3}{16g} \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( \dfrac{2v}{g}\sec\theta \)

Explanation:
1. Take axes along and perpendicular to the incline: perpendicular velocity v, perpendicular acceleration \( g\cos\theta \).
2. \( T = 2v/(g\cos\theta) = (2v/g)\sec\theta \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (b) \( 20\sqrt2\text{ m s}^{-1} \)

Explanation:
1. At B the speed must give range 40 m at 45°: \( v_B^2/g = 40 \), so \( v_B^2 = 400 \). Height of B is \( 20\sqrt2\sin45^\circ = 20 \) m.
2. Energy: \( v_0^2 = v_B^2 + 2g(20) = 400+400 = 800 \), so \( v_0 = 20\sqrt2 \) m/s.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( \dfrac{a^2}{4b},\ \tan^{-1}a \)

Explanation:
1. Slope at the origin is \( dy/dx = a = \tan\theta \). Maximum of y occurs at \( x = a/2b \).
2. \( y_{max} = a\cdot\dfrac{a}{2b} - b\dfrac{a^2}{4b^2} = \dfrac{a^2}{4b} \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (b) \( 2\sqrt3h \)

Explanation:
1. \( R = u^2\sin2\theta/g = 4gh\sin120^\circ/g = 4h\cdot\dfrac{\sqrt3}{2} \).
2. \( R = 2\sqrt3\,h \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( 2\sqrt{h/g} \)

Explanation:
1. \( u_y = 2\sqrt{gh}\sin60^\circ = \sqrt{3gh} \). Solve \( h = u_yt - \tfrac12gt^2 \): \( t = \dfrac{\sqrt{3gh}\pm\sqrt{gh}}{g} \).
2. Time above h \( = t_2 - t_1 = 2\sqrt{gh}/g = 2\sqrt{h/g} \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (d) \( 3h/2 \)

Explanation:
1. \( H = u^2\sin^260^\circ/2g = 4gh\cdot\tfrac34/2g \).
2. \( H = 3h/2 \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (d) \( 20\sqrt3\text{ m s}^{-1} \)

Explanation:
1. The top is reached at t = 3 s, so \( u_y = 30 \) m/s and at t = 2 s \( v_y = 10 \) m/s. Then \( \tan30^\circ = 10/v_x \) gives \( v_x = 10\sqrt3 \).
2. \( u = \sqrt{30^2+(10\sqrt3)^2} = \sqrt{1200} = 20\sqrt3 \) m/s.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( 60^\circ \)

Explanation:
1. \( u_y = 30 \) m/s and \( u_x = 10\sqrt3 \) m/s.
2. \( \tan\theta = 30/(10\sqrt3) = \sqrt3 \), so \( \theta = 60^\circ \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (b) \( y = 2x - 5x^2 \)

Explanation:
1. \( y = x\tan\theta - \dfrac{gx^2}{2u_x^2} \) with \( \tan\theta = 2 \) and \( u_x = 1 \).
2. \( y = 2x - 5x^2 \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( r_1:r_2 \)

Explanation:
1. \( a = 4\pi^2r/t^2 \) for each car with the same t.
2. So \( a_1:a_2 = r_1:r_2 \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (d) \( 20\text{ m} \)

Explanation:
1. Vertical throw: \( u^2 = 2gH = 200 \).
2. Maximum range at 45° is \( u^2/g = 20 \) m.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (d) \( \dfrac{\sqrt3}{16}\dfrac{mv^3}{g} \)

Explanation:
1. At the top \( v_x = v\cos30^\circ \) and \( H = v^2\sin^230^\circ/2g = v^2/8g \).
2. \( L = mv_xH = m\cdot\dfrac{\sqrt3v}{2}\cdot\dfrac{v^2}{8g} = \dfrac{\sqrt3mv^3}{16g} \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

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

Explanation:
1. The maximum range is \( R = v^2/g \).
2. The bullets cover a circle of radius R: area \( \pi v^4/g^2 \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (d) fourth

Explanation:
1. At the nth step edge: \( x = 0.2n = 1.8t \) and the fall \( y = 5t^2 = 5(0.2n/1.8)^2 = 0.0617n^2 \).
2. The ball is below the nth step when \( 0.0617n^2 \ge 0.2n \), i.e. \( n\ge3.24 \); the first such step is the fourth.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) \( \dfrac{R}{1-k^2} \)

Explanation:
1. Energy: \( \tfrac12m k^2v_e^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{r} \) with \( v_e^2 = 2GM/R \).
2. \( \dfrac{GM}{R}(k^2-1) = -\dfrac{GM}{r} \), so \( r = \dfrac{R}{1-k^2} \).

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (a) \( 10\sqrt5\text{ m s}^{-1} \)

Explanation:
1. Time to top = 2 s. Horizontal displacement \( = 20\times2 = 40 \) m; height \( = 20^2/20 = 20 \) m.
2. Displacement \( = \sqrt{40^2+20^2} = 20\sqrt5 \) m, so average velocity \( = 20\sqrt5/2 = 10\sqrt5 \) m/s.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.

Answer: (c) 21%

Explanation:
1. \( H\propto u^2 \).
2. \( (1.1)^2 = 1.21 \), an increase of 21%.

Teacher's Note:
1. Resolve the motion into horizontal and vertical components first.
2. Use g = 10 m s⁻² unless stated otherwise.