JEE Advanced Physics Non-Uniform Motion PYQ

Q 1 :

The velocity-displacement graph of a particle moving along a straight line is shown [2005]

The most suitable acceleration-displacement graph will be

1

2

3

4

(1)

$The equation for the given v - x graph is$

$v = - \frac{v_{0}}{x_{0}} x + v_{0} ...(i)$

$\frac{d v}{d x} = - \frac{v_{0}}{x_{0}}$

$∴$ $a = v \frac{d v}{d x} = - \frac{v_{0}}{x_{0}} \times v = - \frac{v_{0}}{x_{0}} [- \frac{v_{0}}{x_{0}} x + v_{0}] from (i)$

$\Rightarrow a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}} ...(ii)$

$On comparing equation (ii) with equation of a straight line$

$y = m x + c$

$we get m = \frac{v_{0}^{2}}{x_{0}^{2}} = +ve,$

$i.e., \tan θ = + ve, i.e., θ is acute.$

$Also c = - \frac{v_{0}^{2}}{x_{0}^{2}}$ ,

$i.e., the y-intercept is negative. Hence graph (a) correctly depicts corresponding a - x graph.$

Q 2 :

A body starts from rest at time $t = 0$ , the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be [2004]

110 m/s
55 m/s
650 m/s
550 m/s

1

2

3

4

(2)

$Change in velocity = area under the acceleration-time graph$

$= \frac{1}{2} \times 10 \times 11 = 55 m/s$

$Since, initial velocity is zero, final velocity i.e., maximum velocity is 55 m/s.$

Q 3 :

A rocket is moving in a gravity free space with a constant acceleration of $2 {m/s}^{2}$ along $+ x$ direction (see figure). The length of a chamber inside the rocket is $4 m$ . A ball is thrown from the left end of the chamber in $+ x$ direction with a speed of $0.3 m/s$ relative to the rocket. At the same time, another ball is thrown in $- x$ direction with a speed of $0.2 m/s$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is [2014]

(8)

$For ball A$

$u_{1} = 0.3 {ms}^{- 1}, a_{1} = - 2 {ms}^{- 2}, s_{1} = x, t_{1} = t$

$∴$ $s_{1} = u_{1} t_{1} + \frac{1}{2} a_{1} t_{1}^{2}$

$x = 0.3 t - t^{2} ...(i)$

$For ball B$

$u_{2} = 0.2 {ms}^{- 1}, a_{2} = 2 {ms}^{- 2}, s_{2} = 4 - x, t_{2} = t$

$∴$ $s_{2} = u_{2} t_{2} + \frac{1}{2} a_{2} t_{2}^{2}$

$4 - x = 0.2 t + t^{2} ...(ii)$

$Adding eq. (i) and (ii):$

$4 = 0.5 t$ $∴$ $t = \frac{4}{0.5} = 8 s$

Note: Particle 'A' will never move as $a_{1} = - 2 {m/s}^{2}$ is much larger than its initial velocity of $0.2 m/s$ . But to match the official answer, we solved as shown above.

Topic Question Set

Q 1 :

The velocity-displacement graph of a particle moving along a straight line is shown [2005]

The most suitable acceleration-displacement graph will be

Q 2 :

A body starts from rest at time $t = 0$ , the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be [2004]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : The velocity-displacement graph of a particle moving along a straight line is shown [2005] The most suitable acceleration-displacement graph will be

Q 2 : A body starts from rest at time t=0, the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be [2004]

Want Us to Email you About Special Offers & Updates?

Q 1 :

The velocity-displacement graph of a particle moving along a straight line is shown [2005]

The most suitable acceleration-displacement graph will be

Q 2 :

A body starts from rest at time $t = 0$ , the acceleration time graph is shown in the figure. The maximum velocity attained by the body will be [2004]