Laws of Motion MCQs for NEET — Physics Questions with Answers

For a bullet to be stopped, it must lose all its velocity. If it loses 1/20th of its velocity after passing through one plank, then to lose its entire velocity, it needs to pass through 20 planks. Therefore, the least number of planks required to stop the bullet is 11.

$v=4t^3-2t$(given) ∴ $a=\frac{dv}{dt}=12t^2-2$

and $x={\int }_0^{t}vdt={\int }_0^t(4t^3-2t)dt=t^4-t^2$

When particle is at 2m from the origin $t^4-t^2=2$

⇒ $t^4-t^2-2=0$ ⇒ $(t^2-2)(t^2+1)=0$ ⇒ $t=\sqrt{2}\sec$

Acceleration at $t=\sqrt{2}\sec$ given by,

$a=12t^2-2=12\times 2-2$ = $22m/s^2$  

$$\begin{gathered} \frac{dt}{dx}=2\alpha x+\beta \Rightarrow v=\frac{1}{2\alpha x+\beta } \\ Let, 2\alpha x+\beta =p\Rightarrow \frac{dp}{dx}=2\alpha \\ \Rightarrow v=\frac{1}{p} \\ \Rightarrow \frac{dv}{dp}=-\frac{1}{p^2}=\frac{-1}{{(2\alpha x+\beta )}^2} \\ Now, \frac{dv}{dx}=\frac{dv}{dp}\times \frac{dp}{dx}=\frac{-2\alpha }{{(2\alpha x+\beta )}^2} \end{gathered}$$

$\because a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}$

$a=v\frac{dv}{dx}=\frac{-v.2\alpha }{{(2\alpha x+\beta )}^2}=-v.2\alpha \times \frac{1}{{(2\alpha x+\beta )}^2}=-2\alpha .v.v^2=-2\alpha v^3$

∴ Retardation $=2\alpha v^3$  

Let $u_1,u_2,u_3$ and u₄ be velocities at time $t=0,t_1,(t_1+t_2)$ and $(t_1+t_2+t_3)$ respectively and acceleration is a then $v_1=\frac{u_1+u_2}{2},v_2=\frac{u_2+u_3}{2}\text{and\hspace{0.17em}}{v}_3=\frac{u_3+u_4}{2}$

Also $u_2=u_1+a{t}_1,u_3=u_1+a(t_1+t_2)$

and $u_4=u_1+a(t_1+t_2+t_3)$

By solving, we get $\frac{v_1-v_2}{v_2-v_3}=\frac{(t_1+t_2)}{(t_2+t_3)}$  

$$\begin{gathered} Let initial velocity of car be u and it stops after covering dis\tan ce S \\ By 3rd equation of motion \\ {v}^2 =u^2 -2aS \\ Where v =final velocity \\ To stop car , v=0 \\ Putting in above equation, \\ S= \frac{u^2}{2a} \\ where a is retardation and s is the stopping dis\tan ce \end{gathered}$$

$S\propto u^2$.

For constant retardation a, if u becomes 3 times then S will become 9 times i.e. $9\times 20=180m$  

where S is the stopping distance.

$S\propto u^2$ 

$\frac{S_1}{S_2}={\left(\frac{u_1}{u_2}\right)}^2$ ⇒ $\frac{2}{S_2}=\frac{1}{4}\Rightarrow S_2=8m$ 

Using relative motion concept,$$\begin{gathered} {\overrightarrow{a}}_{bolt,car} = {\overrightarrow{a}}_{bolt} - {\overrightarrow{a}}_{car} \\ = 9.8 - (-1.2) \\ = 11 m/s^2 \\ Also initial velocity of bolt wrt car u = 0 \end{gathered}$$

$t=\sqrt{\frac{2h}{(g+a)}}=\sqrt{\frac{2\times 2.7}{(9.8+1.2)}}=\sqrt{\frac{5.4}{11}}=\sqrt{0.49}=0.7\sec$

As u = 0 and lift is moving upward with acceleration.

Displacement $x=2t^2+t+5$

Velocity $=\frac{dx}{dt}=4t+1$

Acceleration $=\frac{d^{2}x}{d{t}^2}=4$ i.e. independent of time

Hence acceleration $=4m/s^2$   

$v=u+at=u+\left(\frac{F}{m}\right)t=20+\left(\frac{100}{5}\right)\times 10=220m/s$    

Mass does not affect on maximum height.

$H=\frac{u^2}{2g}\Rightarrow H\propto u^2$, So if velocity is doubled then height will become four times. i.e. $H=20\times 4=80m$