Question

A particle is projected up a large inclined plane from a point $O$ as shown in the figure. The projection velocity has a magnitude of $5.5m{s}^{-1}$ and its direction makes an angle ${37}^{\circ }$ with the inclined plane. The inclination of the plane is also ${37}^{\circ }$. The inclined plane starts moving towards left with an acceleration $a_0=5m{s}^{-2}$ at the moment the particle is projected. If the particle strikes the inclined plane at a point $P$, choose the correct statements.

• A. The time taken by the particle to move from $O$ to $P$ is $0.6s$
• B. The time taken by the particle to move from $O$ to $P$ is $0.4s$
• C. The displacement of the particle w.r.t. the inclined plane is $2.56m$
• D. The displacement of the particle w.r.t. the inclined plane is $2.28m$

Answer 1

Answer: (A), (D)

The correct options are
A The time taken by the particle to move from $O$ to $P$ is $0.6s$
D The displacement of the particle w.r.t. the inclined plane is $2.28m$


w.r.t the wedge the accelerations of the particle are,

$a_x=-g\sin {37}^{\circ }+a_0\cos {37}^{\circ }$
$a_x=-10\times \frac{3}{5}+5\times \frac{4}{5}$
$a_x=-2m{s}^{-2}$
$a_y=-g\cos {37}^{\circ }-a_0\sin {37}^{\circ }$
$a_y=-10\times \frac{4}{5}-5\times \frac{3}{5}$
$a_y=-11m{s}^{-2}$

Using Kinematics equation
$x=5.5\cos {37}^{\circ }t-\frac{1}{2}\left(2\right)t^2=4.4t-t^2$
$y=5.5\sin {37}^{\circ }t-\frac{1}{2}\left(11\right)t^2=3.3t-5.5t^2$

When particle reaches $P$, $y=0,t=T$
$\Rightarrow 3.3T=5.5T^2$
$\Rightarrow t=\frac{3.3}{5.5}=0.6s$

The displacement w.r.t. inclined plane is the range $\left(R\right),x=R\text{at}t=0.6s$.
$R=4.4\left(0.6\right)-(0.6{)}^2$
$R=2.64-0.36$
$R=2.28m$