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NEET
Exam
NEET
Subject
Physics
Chapter
Motion in Two Dimensions

NEET · Physics · Motion in Two Dimensions

A particle is projected up from a point at an angle θ with the...

A particle is projected up from a point at an angle θ\theta with the horizontal direction. At any time tt, if pp is the linear momentum, yy is the vertical displacement, xx is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy KE of the particle is A particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear

Options

  1. Agraph (A)
  2. Bgraph (B)
  3. Cgraph (C)
  4. Dgraph (D)

Correct answer

B. graph (B)

Step-by-step solution

Momentum, p=m⋅v\quad p=m \cdot v A particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear
⇒v=(pm) Kinetic energy, KE=12mv2=12m(p2m2)=12mp2⇒KE∝p2(∵12m= constant )\begin{aligned} & \Rightarrow \quad v=\left(\frac{p}{m}\right) \\ & \text{ Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2 \\ & =\frac{1}{2} m\left(\frac{p^2}{m^2}\right)=\frac{1}{2 m} p^2 \\ & \Rightarrow \quad \mathrm{KE} \propto p^2 \quad\left(\because \frac{1}{2 m}=\text{ constant }\right) \\ & \end{aligned} Hence, the graph between KE and p2p^2 will be linear as shown belowA particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear
Now, kinetic energy KE =12\frac{1}{2} m v^2ThevelocitycomponentatpointThe velocity component at pointP,and, andvy=(usin⁡θ−gt)vx=ucos⁡θ\begin{aligned} & v_y=(u \sin \theta-g t) \\ & v_x=u \cos \theta \end{aligned}ResultantvelocityatpointResultant velocity at pointP,,v→=vyj^+vxi^=(usin⁡θ−gt)j^+ucos⁡θi^∣v→∣=(ucos⁡θ)2+(usin⁡θ−gt)2=u2cos⁡2θ+u2sin⁡2θ+g2t2−2ugtsin⁡θ∴=u2(cos⁡2θ+sin⁡2θ)+g2t2−2ugtsin⁡θKE=12m(u2+g2t2−2ugtsin⁡θ)⇒KE∝t2\begin{aligned} & \overrightarrow{\mathbf{v}}=v_y \hat{\mathbf{j}}+v_x \hat{\mathbf{i}} \\ & =(u \sin \theta-g t) \hat{\mathbf{j}}+u \cos \theta \hat{\mathbf{i}} \\ & |\overrightarrow{\mathbf{v}}|=\sqrt{(u \cos \theta)^2+(u \sin \theta-g t)^2} \\ & =\sqrt{u^2 \cos ^2 \theta+u^2 \sin ^2 \theta+g^2 t^2-2 u g t \sin \theta} \\ & \therefore=\sqrt{u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+g^2 t^2-2 u g t \sin \theta} \\ & \mathrm{KE}=\frac{1}{2} m\left(u^2+g^2 t^2-2 u g t \sin \theta\right) \\ & \Rightarrow \mathrm{KE} \propto t^2 \\ & \end{aligned}Hence,graphwillbeparabolicwithinterceptonHence, graph will be parabolic with intercept ony−axis.Hence,thegraphbetweenKEand-axis. Hence, the graph between KE andtA particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear
Now, in case of heightKE=12mheight \mathrm{KE}=\frac{1}{2} m\left(v^2\right)andandv^2=(u2−2gy)\left(u^2-2 g y\right) ∴KE=12m(u2−2gy)KE=−mgy+12mu2\begin{aligned} \therefore \quad \mathrm{KE} & =\frac{1}{2} m\left(u^2-2 g y\right) \\ \mathrm{KE} & =-m g \mathrm{y}+\frac{1}{2} m u^2 \end{aligned}InterceptonIntercept ony−axis=12mu2-axis =\frac{1}{2} m u^2 Now, KE=12mv2KE=12m(xt)2\begin{aligned} \mathrm{KE} & =\frac{1}{2} m v^2 \\ \mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2 \end{aligned}A particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear
Now, KE=12mv2KE=12m(xt)2\begin{aligned} \mathrm{KE} & =\frac{1}{2} m v^2 \\ \mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2 \end{aligned}A particle is projected up from a point at an angle θ with the horizontal direction. At any time t , if p is the linear
KE∝x2\mathrm{KE} \propto x^2. Thus graph between KE and xx will be parabolic.
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