Ball hitting an inclined plane

[terryds]

Homework Statement

http://www.sumoware.com/images/temp/xzafcttptkeoehmo.png [Broken]
An inclined plane is put on a smooth floor. The inclined plane is hit (collided) by an elastic ball moving horizontally before the collision. The ball bounce from the inclined plane and land again right at the point of first collision. If the inclination angle is Θ, determine the ratio between the ball mass and the inclined plane mass

Homework Equations

p_x = p_x'
E = E'

The Attempt at a Solution

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Actually I don't know what to do, so I just draw the condition after collision
http://www.sumoware.com/images/temp/xzknipxthmfhalxb.png [Broken]
Vp is the velocity of inclined plane

p = p
m_b v_b = m_b v_bx' + m_p v_p'

Then, I don't know what to do
Please help me

 

[ehild]

Energy is conserved, as the ball is elastic and the ground is smooth.
The ball reaches the plane at the first collision point again. What does it mean for the horizontal distances traveled by the plane and by the ball? What information you have about the vertical and horizontal velocity components of the ball?

 

[terryds]

What does it mean for the horizontal distances traveled by the plane and by the ball?

Hmm... When the ball reaches the plane at the first collision point, it means that the horizontal distance traveled by the ball is the same as traveled by the plane.
So
p = p
mb vb = mb vbx' + mp vp'
m_b v_b = (m_b + m_p)v'_x

What information you have about the vertical and horizontal velocity components of the ball?

The vertical component of the ball is v'_by = v cos Θ
Horizontal component is v'_bx = v sin Θ

Energy is conserved, as the ball is elastic and the ground is smooth.

(1/2)mv_b² = (1/2)mv_b'x² + (1/2)mv_p'x²
(1/2)mv_b² = mv_b'x²
v_b² = 2v_b'x²
v_b² = 2(v_b sin Θ)²
v_b = √(2(v_b sin Θ)²)
Oops.. It seems like a dead-end to me.. Please help me

 

[ehild]

Hmm... When the ball reaches the plane at the first collision point, it means that the horizontal distance traveled by the ball is the same as traveled by the plane.
So
p = p
mb vb = mb vbx' + mp vp'
m_b v_b = (m_b + m_p)v'_x

Correct.

The vertical component of the ball is v'_by = v cos Θ
Horizontal component is v'_bx = v sin Θ

Why?

 

[terryds]

Correct.

Why?

Because if theta increases, the y component of the velocity of ball after collision decreases, right ? So, it's cosine function, right ?
And if theta increases, the x component of the velocity of ball after collision increases, right ? So, it's sinus function, right ?
But, I don't know what it will be if the 'launch' angle has nothing to do with the angle theta

 

[haruspex]

Because if theta increases, the y component of the velocity of ball after collision decreases, right ? So, it's cosine function, right ?
And if theta increases, the x component of the velocity of ball after collision increases, right ? So, it's sinus function, right ?
But, I don't know what it will be if the 'launch' angle has nothing to do with the angle theta

There are trig functions besides cos and sine.
Let the impulse the block delivers to the ball be J. Which way does it point? What is its component in the x direction?

 

[terryds]

http://www.sumoware.com/images/temp/xzaciotcpqkxexdr.png [Broken]
J = mg cos Θ Δt
J_x = m v'_bx - m v_b

But, what is v'_bx ?

 

[ehild]

The plane can exert force only in the perpendicular direction, as you drew, but the impulse is not connected to gravity. The parallel (with the plane) component of the initial velocity of the ball does not change during the collision. Only the perpendicular component changes. Conservation of energy provides a relation among the speed of the slope V_p, initial speed of the ball v_b, and the perpendicular component after collision, v_m.

Both the perpendicular and parallel components of the velocity after collision have horizontal components, depending on the angle theta. You can get an expression for v'_bx in terms of theta.
You know that v'bx=V_p, and conservation of momentum yields V_p=m_bv_b/(m_b+m_p). There are enough equation to get the mass ratio in terms of theta.

 

[haruspex]

J = mg cos Θ Δt

As ehild says, g doesn't enter into it. Δt is very small.
Deduce J from J_x (simple trig), and deduce J_y from that.
Or use ehild's observation that "The parallel (with the plane) component of the initial velocity of the ball does not change during the collision", which I was trying to avoid assuming. The two methods amount to the same.