Physics Q & A

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

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

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

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 $\vec{r(t)}$