JEE Physics Motion In A Plane MCQs Set 26

Answer: (b) \( R^2=16h_1h_2 \)

Explanation:
1. Same range means the angles are \( \theta \) and \( 90^\circ-\theta \). \( h_1=\dfrac{u^2\sin^2\theta}{2g} \), \( h_2=\dfrac{u^2\cos^2\theta}{2g} \).
2. \( h_1h_2=\dfrac{u^4\sin^2\theta\cos^2\theta}{4g^2}=\dfrac{R^2}{16} \) since \( R=\dfrac{u^2\cdot2\sin\theta\cos\theta}{g} \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (d) \( \dfrac{2R}g \)

Explanation:
1. The two angles are \( \theta \) and \( 90^\circ-\theta \): \( t_1=\dfrac{2u\sin\theta}g \), \( t_2=\dfrac{2u\cos\theta}g \).
2. \( t_1t_2=\dfrac{4u^2\sin\theta\cos\theta}{g^2}=\dfrac{2R}g \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( \theta_0=\cos^{-1}\left(\tfrac1{\sqrt5}\right),\ v_0=\tfrac53\ \text{m s}^{-1} \)

Explanation:
1. Compare with \( y=x\tan\theta_0-\dfrac{gx^2}{2v_0^2\cos^2\theta_0} \): \( \tan\theta_0=2 \), so \( \cos\theta_0=1/\sqrt5 \).
2. \( \dfrac{g}{2v_0^2\cos^2\theta_0}=9\Rightarrow v_0^2=\dfrac{10}{18\cdot(1/5)}=\dfrac{25}9 \), so \( v_0=\tfrac53 \) m s⁻¹.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) 20 cm

Explanation:
1. Range along an incline: \( R=\dfrac{2u^2\sin\theta\cos(\theta+\alpha)}{g\cos^2\alpha} \) with \( \theta \) measured from the plane.
2. \( R=\dfrac{2(4)\sin15^\circ\cos45^\circ}{10\cos^230^\circ}=\dfrac{8(0.2588)(0.7071)}{7.5}=0.195 \) m, close to 20 cm.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (a) 50

Explanation:
1. \( \vec a=\dfrac{d^2\vec r}{dt^2}=30\hat i-40\hat j \).
2. \( |\vec a|=\sqrt{900+1600}=50 \) m s⁻².

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( a+\dfrac{b^2}{3c} \)

Explanation:
1. \( a_{cc}=2b-6ct=0\Rightarrow t=\dfrac{b}{3c} \).
2. \( v=a+2bt-3ct^2=a+\dfrac{2b^2}{3c}-\dfrac{b^2}{3c}=a+\dfrac{b^2}{3c} \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) 2.8 m

Explanation:
1. At t = 1 s: \( v_x=5 \), \( v_y=5\sqrt3-10=-1.34 \), so \( v^2=25+1.8=26.8 \) and \( v=5.18 \) m s⁻¹.
2. Normal component of g: \( a_n=g\,v_x/v=10(5/5.18)=9.66 \) m s⁻².
3. \( R=v^2/a_n=26.8/9.66=2.8 \) m.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) 1 : 16

Explanation:
1. Maximum range \( =u^2/g \); the area covered is a circle \( \pi R_{max}^2\propto u^4 \).
2. Ratio \( =(1/2)^4=1:16 \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

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

Explanation:
1. \( u_x=1,\ u_y=2 \): \( y=\dfrac{u_y}{u_x}x-\dfrac{g}{2u_x^2}x^2 \).
2. \( y=2x-5x^2 \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

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

Explanation:
1. \( dy/dx=0 \Rightarrow x=a/2b \), so \( y_{max}=\dfrac{a^2}{2b}-\dfrac{a^2}{4b}=\dfrac{a^2}{4b} \).
2. The slope at the origin is \( \tan\theta=a \), so \( \theta=\tan^{-1}a \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (a) 5

Explanation:
1. Time of flight \( T=\dfrac{2(10)\sin60^\circ}{10}=\sqrt3 \) s. Horizontal displacement of the ball (ground frame) \( =10\cos60^\circ\cdot T=5\sqrt3=8.66 \) m.
2. The boy (train) moves \( \tfrac12aT^2=1.5a \) m in the same time, and the ball lands 1.15 m ahead of him: \( 8.66-1.5a=1.15 \).
3. \( a\approx5 \) m/s².

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (a) 1 s

Explanation:
1. Horizontal shot: \( x=5\sqrt3\,t_1,\ y=10-5t_1^2 \). Inclined shot: \( x=\tfrac{5\sqrt3}2t_2,\ y=10+\tfrac{15}2t_2-5t_2^2 \).
2. Equal x: \( t_2=2t_1 \). Equal y: \( -5t_1^2=7.5(2t_1)-20t_1^2 \Rightarrow15t_1^2=15t_1\Rightarrow t_1=1 \) s, \( t_2=2 \) s (collision at \( (5\sqrt3,\,5) \) m).
3. The time interval \( =t_2-t_1=1 \) s.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.