## A tennis ball is released from height d and after freely falling on a wooden floor it rebounds and reaches height d/2 ….

Q: A tennis ball is released from height d and after freely falling on a wooden floor it rebounds and reaches height d/2 . The velocity versus height of the ball during its motion may be represented graphically by

Click to See Answer :
Ans: (a)

Sol: When ball is falling down from height d and reaches at height h , then velocity acquired by the ball is

v2= 2g(d-h)    ….(i)

This is downward velocity and is taken as negative .

When h = d , v = 0

When h =0 , v2 = (2 g d)

Hence from eqn(i) , the variation of v & h is parabolic below h axis

## Train A and Train B are running on parallel track in the opposite direction with speed of 36 km/h and 72 km/h respectively ….

Q: Train A and Train B are running on parallel track in the opposite direction with speed of 36 km/h and 72 km/h respectively . A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/h . Speed ( in m/s ) of this person as observed from train B will be close to ( take the distance between the tracks are negligible )

(a) 29.5 m/s

(b) 28.5 m/s

(c) 31.5 m/s

(d) 30.5 m/s

Ans: (a)

Sol: Let train A is moving along x-axis .

Velocity of train A with respect to ground , $\displaystyle \vec{v_A} = 36 \hat{i} \; km/h$

Velocity of train B with respect to ground , $\displaystyle \vec{v_B} = 72 (-\hat{i}) \; km/h$

Velocity of person with respect to train A , $\displaystyle \vec{v_{PA}} = -1.8 \hat{i} \; km/h$

Velocity of person with respect to train B , $\displaystyle \vec{v_{PB}} = \vec{v_P} – \vec{v_B}$

$\displaystyle \vec{v_{PB}} = \vec{v_{PA}} + \vec{v_A} – \vec{v_B}$

$\displaystyle \vec{v_{PB}} = -1.8 \hat{i} + 36 \hat{i} – (-72 \hat{i})$

$\displaystyle \vec{v_{PB}} = 106.2 \hat{i} \; km/h$

$\displaystyle \vec{v_{PB}} = 106.2 \hat{i} \times \frac{5}{18} \; m/s$

= 29.5 m/s

## A particle starts from origin at t= 0 with an initial velocity of 3.0 i m/s and moves in x-y plane with a constant acceleration ….

Q: A particle starts from origin at t= 0 with an initial velocity of $3.0 \hat{i} \; m/s$ and moves in x-y plane with a constant acceleration $\displaystyle (6 \hat{i} + 4 \hat{j}) m/s^2$  . The x-co-ordinate of the particle at the instant when its y-co-ordinate is 32 m is D meter . The value of D is

(a) 50

(b) 40

(c) 32

(d) 60

Ans: (d)

Sol: $S_y = u_y t + \frac{1}{2}a_y t^2$

$32 = 0 + \frac{1}{2}(4) t^2$

$32 = 2 t^2$

t = 4 sec

$S_x = u_x t + \frac{1}{2}a_x t^2$

$D = 3 \times 4 + \frac{1}{2}(6) 4^2$ (since u_x = 3 m/s)

D = 12 + 48 = 60 m

## A particle moves such that its position vector r(t) = cosωt i + sinωt j ; where ω is constant and t is time . Then which of the following statement ….

Q: A particle moves such that its position vector $\displaystyle \vec{r(t)} = cosωt \hat{i} + sinωt \hat{j}$ ; where ω is constant and t is time . Then which of the following statement is true for the velocity $\vec{v(t)$ and acceleration $\vec{a(t)}$ of the particle ?

(a) $\displaystyle \vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$

(b) $\displaystyle \vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from origin .

(c) $\displaystyle \vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$

(d) $\displaystyle \vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin .

Ans: (d)

Sol: $\displaystyle \vec{r(t)} = cosωt \hat{i} + sinωt \hat{j}$

$\displaystyle \vec{v(t)} = \frac{d \vec{r}}{dt} = – ω sin ωt \hat{i} + ω cos ωt \hat{j}$

$\displaystyle \vec{a(t)} = \frac{d \vec{v}}{dt} = – ω^2 cos ωt \hat{i} – ω^2 sin ωt \hat{j}$

$\displaystyle \vec{a(t)} = – ω^2 ( cos ωt \hat{i} + sin ωt \hat{j})$

$\displaystyle \vec{a(t)} = – ω^2 \vec{r(t)}$

Negative sign indicates that acceleration is directed towards origin .

$\displaystyle \vec{v(t)}.\vec{r(t)} = (- ω sin ωt \hat{i} + ω cos ωt \hat{j}).(cosωt \hat{i} + sinωt \hat{j})$

$\displaystyle \vec{v(t)}.\vec{r(t)} = – ω sin ωt \; cosωt + ω sin ωt \; cosωt$

$\displaystyle \vec{v(t)}.\vec{r(t)} = 0$

$\vec{v(t)}$ is perpendicular to $\vec{r(t)}$

## A particle is moving in a circle of radius r with a constant speed v . The change in velocity ….

Q: A particle is moving in a circle of radius r with a constant speed ‘ v ‘ . The change in velocity after the particle has travelled a distance equal to (1/8) of the circumference of the circle is :

(a) Zero

(b) 0.500 v

(c) 0.765 v

(d) 0.125 v

Click to See Answer :
Ans: (c)

Sol: Change in velocity , $\displaystyle \vec{\Delta v} = \vec{v_2} – \vec{v_1}$

$\displaystyle |\vec{\Delta v} |= \sqrt{v^2 + v^2 – 2 v \times v \times cos\theta }$ (since v1 = v2 = v and θ = 2π/8)

$\displaystyle |\vec{\Delta v} |= \sqrt{2 v^2 (1 – cos\theta )}$

$\displaystyle |\vec{\Delta v} | = \sqrt{2 v^2 (2 sin^2(\frac{\theta}{2}))}$

$\displaystyle |\vec{\Delta v} | = 2 v sin(\frac{\theta}{2})$

$\displaystyle |\vec{\Delta v} | = 2 v sin(\frac{\pi}{8})$

= 0.765 v