How Do You Solve Projectile Motion on an Incline?

[ebmaj7]

Homework Statement

Part (a) was fine, managed it okay.

Homework Equations

$v_x = 20\cos30 -9.8\sin35 = 3.32$

$v_y = 20\sin30 -9.8\cos35 = -10$

Part (b) is what is giving me trouble at the moment.

The Attempt at a Solution

$\dfrac{4}{5} = \dfrac{v_b - v_a}{u_a - u_b}$

I really don't know where to go from here.

The plane isn't moving. So if the plane was object 'b' then v_b and u_b would both be 0.

The speed at which the projectile strikes point A (the final velocity) = the initial velocity after the collision has taken place.

You could then solve to find the velocity of the projectile after the collision.

Revising for my Mechanics 3 A level exam, AQA.

 

[anathema]

Hello there! Part (b) is definitely a bit trickier, but don't worry, we can work through it together. Let's start by writing out the equations we know:

- The horizontal component of the projectile's velocity: v_x = 3.32 m/s
- The vertical component of the projectile's velocity: v_y = -10 m/s

Now, let's consider the collision between the projectile and the plane. We know that the horizontal component of the projectile's velocity will not change after the collision since the plane is not moving. So, the horizontal component of the final velocity (after the collision) will also be 3.32 m/s.

For the vertical component, we can use the equation of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the momentum is given by mass times velocity (p = mv).

So, before the collision, the momentum of the projectile is:

p_a = m_a * v_a = 4 kg * 20 m/s = 80 kg*m/s

After the collision, the momentum of the projectile is:

p_b = m_b * v_b = 4 kg * v_b

And the momentum of the plane is:

p_p = m_p * v_p = 0 kg * 0 m/s = 0 kg*m/s

Since the total momentum before the collision is equal to the total momentum after the collision, we can write:

p_a = p_b + p_p

Substituting in the values we know, we get:

80 kg*m/s = 4 kg * v_b + 0 kg*m/s

Solving for v_b, we get:

v_b = 20 m/s

So, the final velocity of the projectile after the collision is:

v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.32 m/s)^2 + (20 m/s)^2} = \sqrt{434.24} = 20.84 m/s

I hope this helps! Good luck on your exam!