## A man is holding an umbrella at angle 30° with vertical with lower end towards himself, which is appropriate angle ….

Q: A man is holding an umbrella at angle 30° with vertical with lower end towards himself, which is appropriate angle to protect him from rain for his horizontal velocity 10 m/s. Then which of the following will be true- (A) rain is falling at angle 30° with vertical, towards the man

(B) rain may be falling at angle 30° with vertical, away from the man

(C) rain is falling vertically

(D) none of these

Ans: (B)

Solution: Let v = velocity of rain ; Hence Possible values of α are -30º < α < 90º

## A ball is thrown horizontally from the top of a tower. What happens to the horizontal component of its velocity ?

Q: A ball is thrown horizontally from the top of a tower. What happens to the horizontal component of its velocity ?

(a)Increases

(b) Decreases

(c) Remains unchanged

(d) First decreases and then increases

Ans: (c)

Solution: As there is No acceleration in horizontal direction hence the horizontal component of its velocity remains unchanged .

## A swimmer is moving in a square course in a river. The flow velocity of the river is 3 m/s and the velocity of swimmer relative to water is 5 m/s….

Q: A swimmer is moving in a square course in a river. The flow velocity of the river is 3 ms-1 and the velocity of swimmer relative to water is 5 ms-1. The total distance travelled by the swimmer is 80 m.  Time taken by swimmer (in second) to complete the course is 22.5 n . Find the value of n. Ans. 9

## Two particle P and Q are moving in gravity free space with their respective velocities V_p = -i + 2j -2k  ms-1 and V_p = -4i + 2j + 3k  ms-1. At t = to , the separation between them is minimum. Find the value of 10 to .

Q: Two particle P and Q are moving in gravity free space with their respective velocities $V_P = -\hat{i} + 2\hat{j} -2\hat{i}$  ms-1 and $V_Q = -4\hat{i} + 2\hat{j} + 3\hat{k}$ ms-1. At t = to , the separation between them is minimum. Find the value of 10 to .

Ans. 5

## A swimmer is approaching the shore with a speed of 5√3 m/s  . At the instant when it is at a distance of 30√3 m from the shore,…he value of x.

Q: A swimmer is approaching the shore with a speed of 5√3 m/s  . At the instant when it is at a distance of 30√3 m from the shore, he projects a ball at an elevation of 30° for it to just reach the shore. The speed of stone relative to swimmer is 5x ms-1. Find the value of x.

Ans. 4

## Assume ground as XY-plane and vertically upward as Z-axis. The velocity of an aeroplane is 3i + 4 j + 5 k  . A car is moving on ground as such aeroplane is always above the car. At t = 0, both starts from origin of coordinate system. Choose the correct statement(s).

Q: Assume ground as XY-plane and vertically upward as Z-axis. The velocity of an aeroplane is $3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ . A car is moving on ground as such aeroplane is always above the car. At t = 0, both starts from origin of coordinate system. Choose the correct statement(s).

(a) The velocity of car is $3 \hat{i} + 4 \hat{j}$

(b) The equation of path of car is $y = \frac{4}{3}x$

(c) The velocity of aeroplane seen by an observer in car is $5\hat{k}$

(d) None of the above

Ans: (a),(b),(c)

## A train is going fast a station with a speed of 5 ms-1 towards positive X-axis. A particle resting on the floor of train is giving an acceleration of 4 ms-2 in a direction perpendicular to train’s motion. The trajectory of the particle seen by an observer on the platform is

Q: A train is going fast a station with a speed of 5 ms-1 towards positive X-axis. A particle resting on the floor of train is giving an acceleration of 4 ms-2 in a direction perpendicular to train’s motion. The trajectory of the particle seen by an observer on the platform is

(a) circle

(b) ellipse

(c) ellipsoid

(d) hyperbola

Ans: (c)

## On a frictionless horizontal surface, assumed to be the XY-plane, a small trolley A is moving along a straight line parallel to the Y-axis with a constant velocity of  ( √3-1) m/s …

Q: On a frictionless horizontal surface, assumed to be the XY-plane, a small trolley A is moving along a straight line parallel to the Y-axis (as shown in figure) with a constant velocity of  ( √3-1) m/s  . At a particular instant, when the line OA makes an angle of 45° with the X-axis, ball is thrown along the surface from the origin O. Its velocity makes an angle φ with the X-axis and it hits the trolley. (a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity vector of the ball with the X axis in this frame.

(b) Find the speed of the ball with respect of the surface, it φ = 4θ/3.

Solution :  (a) Let A represents Trolley and B represent Ball

Relative velocity of Ball with respect to A (vAB) should be along OA for the ball to hit the trolley . Hence vBA will make an angle of 45° with +ve X-axis .

(b) Suppose v = absolute velocity of the ball .

φ = 4θ/3 = φ = 4/3(45°) = 60° with X-axis .

$\displaystyle \vec{v_B} = v cos\theta \hat{i} + v sin\theta \hat{j}$

$\displaystyle \vec{v_B} = \frac{v}{2}\hat{i} + \frac{\sqrt{3}v}{2}\hat{j}$

$\displaystyle \vec{v_A} = (\sqrt{3}-1) \hat{j}$

$\displaystyle \vec{v_{BA}} = \frac{v}{2}\hat{i} + (\frac{\sqrt{3}v}{2} – \sqrt{3}+ 1) \hat{j}$

Since vBA is at 45°

$\displaystyle \frac{v}{2} = \frac{\sqrt{3}v}{2} – \sqrt{3}+ 1$

On solving v = 2 m/s

## An object A is kept fixed at the point x =3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the + x-direction ….

Q: An object A is kept fixed at the point x =3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the + x-direction with an acceleration 1.5 ms-2. At the same instant, a stone is projected from the origin with a velocity u as shown in below figure. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45° to the horizontal. At the motions are in xy-plane. Find u and the time after which the stone hits the object. (Given, g = 10 ms-2)

Solution : Let t be the time after which the stone hits the object and θ be the angle which the velocity vector u makes with horizontal . According to question , we have following three conditions .

Vertical displacement of stone is 1.25 m

1.25 = (u sinθ ) t – (1/2)g t2

1.25 = (u sinθ ) t – 5 t2

(u sinθ ) t = 1.25 + 5 t2 …(i)

Horizontal displacement of stone = 3 + displacement of object A

(u cosθ ) t = 3 + (1/2)a t2 ; Where a = 1.5 m/s2

(u cosθ ) t = 3 + 0.75 t2 …(ii)

Horizontal component of velocity (of Stone) = Vertical component
(because velocity vector is inclined at 45° with horizontal )

u cosθ = g t – u sinθ …(iii)
(As stone is in its downward direction)
Therefore g t > u sinθ . In upward motion u sinθ > g t

Multiplying Equation (iii) with t ,

u cosθ t + u sinθ t = g t2

u cosθ t + u sinθ t = 10 t2 …(iv)

Now Equation (iv) , (ii) & (i) gives

4.25 t2 – 4.25 = 0

t= 1 sec

putting t = 1 sec in Equation (i) & (ii) we get

u sinθ = 6.25 m/s or , uy = 6.25 m/s

u cosθ = 3.75 m/s or , ux = 3.75 m/s

$\displaystyle \vec{u} = u_x \hat{i} + u_y \hat{j}$

$\displaystyle \vec{u} = 3.75 \hat{i} + 6.25 \hat{j}$ m/s

## A large heavy box is sliding without friction down a smooth plane of inclination  θ . From a point P on the bottom of the box, a particle is projected inside the box…

Q: A large heavy box is sliding without friction down a smooth plane of inclination  θ . From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle α with the bottom as shown in the figure. (a) Find the distance along the bottom of the box between the point of projection P and the point Q, where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance)

(b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, then find the speed of the box with respect to the ground at the instant when the particle was projected.

Solution : Acceleration of particle with respect to block = Acceleration of particle – Acceleration of block

$\displaystyle = ( g sin\theta \hat{i} + g cos\theta \hat{j} ) – ( g sin\theta ) \hat{i}$

$\displaystyle = g cos\theta \hat{j}$

Now motion of particle with respect to block will be a projectile as shown . The only difference is g will be replaced by g cosθ

Range $\displaystyle R = \frac{u^2 sin2\alpha}{g cos\theta}$

$\displaystyle PQ = \frac{u^2 sin2\alpha}{g cos\theta}$