At time t = 0 a small ball is projected from point A with a velocity of 60 m/s at 60° angle with horizontal…

Q: At time t = 0 a small ball is projected from point A with a velocity of 60 m/s at 60° angle with horizontal. Neglecting atmospheric resistance the two times t1 and t2 when the velocity of the ball makes an angle of 45° with the horizontal x-axis .

(a) 2.19 s, 8.2 s

(b) 3.19 s, 9.2 s

(c) 2.12 s, 8.2 s

(d) 2.19 s, 10.2 s

Ans: (a)

Sol: \displaystyle v_x = u_x = ucos\theta

\displaystyle v_y = usin\theta - gt

\displaystyle tan\alpha = \frac{v_y}{v_x}

\displaystyle tan\alpha = \frac{usin\theta - gt}{ucos\theta}

\displaystyle \pm 1 = \frac{usin\theta - gt}{ucos\theta}

Where , α = ±45° , θ = 60°

on putting α = ±45° , θ = 60° , get two values of t

A body is projected obliquely from the ground such that its horizontal range is maximum. If the change in its linear momentum…

Q: A body is projected obliquely from the ground such that its horizontal range is maximum. If the change in its linear momentum, as it moves from half the maximum height to maximum height, is P, the change in its linear momentum as it travels from the point of projection to the landing point on the ground will be

(a) P

(b) √2 P

(c) 2P

(d) 2√2 P

Ans: (d)

Sol: Maximum height attained is

\displaystyle H = \frac{u^2 sin^2 \theta}{2 g}

Hence , Half of the maximum height is

\displaystyle \frac{1}{2}H = \frac{u^2 sin^2 \theta}{4 g}

From half of maximum height to maximum height :

\displaystyle v_y^2 = u_y^2 + 2a_y S_y

\displaystyle v_y^2 = (usin\theta)^2 - 2 g (\frac{u^2 sin^2 \theta}{4 g})

\displaystyle v_y = \frac{u sin\theta}{\sqrt 2}

Change in momentum from half of maximum height to maximum height :

\displaystyle P = m\frac{u sin\theta}{\sqrt 2}

Required change in momentum

P’ = -2 m u sinθ

P’ = -2(√2 P)

P’ = – 2√2 P

Two persons P and Q crosses the river starting from point A on one side to exactly opposite point B on the other bank …..

Q. Two persons P and Q crosses the river starting from point A on one side to exactly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The person Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each person is 5 kmph w.r.t river. If the two persons reach the point B in the same time, then the speed of walk of Q is.

(a) 13 kmph

(b) 12 kmph

(c) 14 kmph

(d) 11 kmph

Ans:(b)

Sol: Let d = width of river

Time taken to cross the river in shortest path by person P is

\displaystyle t_p = \frac{d}{\sqrt{5^2-3^2}} = \frac{d}{4}

Time taken to cross the river in shortest time by person Q is

\displaystyle t_Q = \frac{d}{5}

According to question

\displaystyle t_p = t_Q + \Delta t ; Where Δt = time taken by person to come to the point B.

If x = walking distance ,

\displaystyle \Delta t = \frac{x}{v_{walk}} ; Whre vwalk = Walking Speed of man Q

\displaystyle \frac{d}{4} = \frac{d}{5} + (v_r\frac{d}{v_Q})\frac{1}{v_{walk}}

\displaystyle \frac{1}{4} = \frac{1}{5} + \frac{3}{5}\frac{1}{v_{walk}}

\displaystyle \frac{1}{4}-\frac{1}{5} = \frac{3}{5}\frac{1}{v_{walk}}

\displaystyle \frac{1}{20} = \frac{3}{5}\frac{1}{v_{walk}}

\displaystyle \frac{1}{4} = \frac{3}{v_{walk}}

vwalk = 12 kmph

A swimmer crosses a flowing stream of width ‘d’ to and fro normal to the flow of the river in time t1. The time taken ….

Q. A swimmer crosses a flowing stream of width ‘d’ to and fro normal to the flow of the river in time t1. The time taken to cover the same distance up and down the stream is t2. If t3 is the time the swimmer would take to swim a distance 2d in still water, then relation between t1, t2 & t3.

(a) \displaystyle t_1 = \sqrt{t_2 t_3}

(b) \displaystyle t_1 = \sqrt{t_2 / t_3}

(c) \displaystyle t_1 = \sqrt{t_3 / t_2}

(d) \displaystyle t_1 = t_3   t_2

Ans: (a)

Sol: Let u = velocity of swimmer
v = velocity of river

\displaystyle t_1 = 2 (\frac{d}{\sqrt{u^2-v^2}} ) …(i)

\displaystyle t_2 = \frac{d}{u+v}+\frac{d}{u-v} …(ii)

\displaystyle t_3 = \frac{2 d}{u} …(iii)

On multiplying (ii) and (iii) we get

\displaystyle t_2 \times t_3 =  t_1^2

\displaystyle  t_1 = \sqrt{t_2 t_3}

Two stones are projected from the top of a tower in opposite direction, with the same velocity v but at 30° & 60° with….

Q. Two stones are projected from the top of a tower in opposite direction, with the same velocity v but at 30° & 60° with horizontal respectively. The relative velocity of first stone relative to second stone is

(a) 2v

(b) √2 v

(c) 2 v/√3

(d) v/√2

Ans: (b)

Sol:

\displaystyle \vec{v_1} = v cos30\hat{i} + v sin30\hat{j}

\displaystyle \vec{v_2} = v cos60\hat{(-i)} + v sin60\hat{j}

\displaystyle \vec{v_{12}} = \vec{v_1}-\vec{v_2}

\displaystyle  = (v cos30 + v cos60)\hat{i} + (v sin30 - v sin60)\hat{j}

\displaystyle  = v[(\frac{\sqrt{3}}{2} +\frac{1}{2})\hat{i} + (\frac{1}{2} -\frac{\sqrt{3}}{2})\hat{j}]

\displaystyle  = v[ \frac{\sqrt{3}+1}{2}\hat{i} - \frac{\sqrt{3}-1}{2}\hat{j}]

\displaystyle v_{12} = \sqrt{2} v