Projectile on a inclined plane

Classical Mechanics Level 4

A particle is projected from the bottom of a inclined plane at an angle $\theta$ with the surface of incline. The incline has an inclination of 1 i.e. angle of incline is 45 degrees. For what value of $\theta$ will the particle retrace its path after collision with the plane.

*Note *

the collision is an elastic collision.
the answer is of the form $\theta$ = $cot^{-1}x$ Find $x$

The answer is 2.

4 solutions

Himanshu Arora
Jun 19, 2014

Let's do this geometrically. So let's take the line $x=y$ and a parabola passing through origin and some point on x axis, whose generic equation is given by $y=-kx(x-a)$ where k and a are positive constants. For retrace of the path in elastic collision, the particle must hit the plane normally. That is when $x=y$ the parabola must be having a tangent of slope $-1$ . Solve these two and get the slope at origin to be 3. And then we'll get the $\tan\theta= 1/2$ . Hence the answer $\boxed{2}$

Hey could u explain me yr solution , I got the answer by physics considering x and y axis perpendicular and parallel to inclined plane. But yr way is much more intriguing.

Vivek Chawla - 6 years, 10 months ago

superb ! the coordinate geometry always works...but i solved it using more of physics by considering the x axis along the incline and y perpendicular to it, then writing the velocity vector and acc. due to gravity vector ..calculating the time of flight and then the velocity with which it strikes the incline and then equating the x component to zero :)

Akshit Vashishth - 6 years, 11 months ago

it is difficult to understand ur ans.

prashant goyal - 6 years, 10 months ago

I couldn't believe that solution. Please explain careful

Mitesh agarwal - 5 years, 4 months ago

Flávio Luis Schneider Júnior
Jan 4, 2015

Let's change our perspective a little bit. Instead of considering that the X and Y plane are in our perspective, we should look at it considering that the X axis is actually the hypotenuse of our inclined plane and decompose gravity along the new X and Y axis, studying the motion as two independent uniformly accelerated motion problem set!

So, decomposing the gravity, we should have a down-oriented g.cos45 acceleration and a left-oriented g.sen 45 acceleration, considering that the particle is launching from (0,0).

To have its path retraced, the particle should hit the plane (or the "floor" in our new perspective) normally, and to do so, it must have its velocity on the X axis reduced to 0, so, this would take (considering the initial velocity equals to "V")

Vx = Vox - g.sin45.t

0 = V.cos(theta) - g.sin45.t

t = [V.cos(theta)]/g.sin45 (eq. I)

So, if it takes "t" to the particle to hit the plane, it would take t/2 to it reaches its maximum height, therefore, its moment where the velocity on the Y axis is zero. So, we have:

Vy = Voy - g.cos45.t/2

0 = V.sin(theta) - g.cos45.t/2

t = [2Vsin(theta)]/g.cos45 (eq. II)

Equaling equations I and II, we have:

[V.cos(theta)]/g.sin45 = [2Vsin(theta)]/g.cos45

It's easy to see then, that:

tan(theta) = 1/2 Therefore, cot(theta) = 2 and theta = cot^(-1) 2

x = 2

Jaikirat Sandhu
Jan 12, 2015

For retracing path, the particle must strike it perpendicular to the incline, so that all velocity is restored in that direction.

Yogesh Ghadge
Aug 28, 2014

the ball will trace the same path when it reaches its maximum height . the incline is at 45 degrees therefore h=R taking equation of Theta = tan^-1(4H/R)= tan^-1(4) =75 theta to the incline=75-45=30 tan30=1/srt3 cot=1/tan cot=srt3=1.77=2 due to errors in angle

Projectile on a inclined plane

The answer is 2.

4 solutions

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