# Spinning ball colliding inclined plane [closed]

[![Solution][2]][2]

"A solid uniform ball of mass M and radius R collides elastically with a rough fixed with a rough inclined surface as shown. Velocity and angular velocity of the ball just before collision are $$v_0$$ and $$\omega_0=\frac{5v_0}{R}$$. Coefficient of friction is 0.75. Find the velocity and the angular velocity just after collision."

How do I approach this problem?

• Please check out our check-my-work policy – noah Feb 22 at 9:49
• All forces are finite. Therefore, the 'just after' should be no difference from 'just before'. – ytlu Feb 22 at 11:33
• @ytlu I’m sure there’s a change in velocities – Boy Feb 22 at 14:00
• @Boy I see it is a colliision problem, My mistake. – ytlu Feb 22 at 14:05
• For the collision to be elastic the friction must be static. In that case, the coefficient gives only the maximum allowed friction force. If that maximum is exceeded, the collision is no longer elastic. That coefficient is useless. – R.W. Bird Feb 22 at 14:31

STEP 1 : Make the components of the velocity, along the incline perpendicular to it. those will be: along incline = $$\frac{3v_{o}}{5}$$ and perpendicular velocity: $$\frac{4v_{o}}{5}$$

STEP 2: Since it is colliding elastically, e=1, so velocity just after collision perpendicular to the incline will be $$\frac{4v_{o}}{5}$$ but opposite to the original direction

STEP 3 Change in momentum = $$\left|2m\cdot\frac{4}{5}v_{o}\right|$$ = $$\int_{ }^{ }Ndt$$

STEP 4 Now impulse due to friction (along incline, of course) will be $$μ\int_{ }^{ }Ndt$$

Now try to solve!

• Thank you I got it – Boy Feb 22 at 14:37
• Why is the normal not constant though? – Boy Feb 22 at 14:37
• I'm thinking that an e=1 applies only to a 1D, non-spinning elastic collision (and that elastic implies conservation of total kinetic energy). – R.W. Bird Feb 22 at 14:55
• @R.W. Bird not sure about that one, have to look up the definition for this. – John Wick Feb 22 at 16:03
• $\mu$ should not enter into the solution. If the collision is elastic, there is no sliding friction, by definition. – Ben51 Feb 22 at 17:14

Taking all the forces in the x direction:

$$\sum{F_{netx}}=mg\sin(37^\circ)-F_{friction}=\frac{3mg}{5}-\mu N=\frac{3mg}{5}-\frac{3mg}{5}=0$$

Hence momentum is conserved in the x direction.

$$\begin{bmatrix} -\frac{3v_0}{5} \cr -\frac{4v_0}{5} \end{bmatrix}\rightarrow\begin{bmatrix} -\frac{3v_0}{5} \cr \frac{4v_0}{5} \end{bmatrix}$$

Impulse along y axis;

$$N\Delta t=\Delta p=\frac{8mv_0}{5}\Rightarrow\Delta t=\frac{8mv_0}{5N}$$

Now for the angular velocity, about the centre of mass about the solid ball

$$\sum{\tau}=r\mu N=I\alpha \Rightarrow \alpha = -\frac{R \mu N}{I}$$

$$\Delta \omega=\alpha \Delta t=-\frac{R\mu 8mv_0}{5I}=-\frac{R\frac{3}{4} 8mv_0}{5\frac{2mR^2}{5}}=-\frac{3v_0}{R}$$ $$\Rightarrow \omega=\frac{2v_0}{R}$$

• This is how I approached the problem. – Boy Feb 22 at 10:47
• There is no reason to assume that the friction force is equal to the component of gravity. – R.W. Bird Feb 22 at 15:19
• Yes okay I got it – Boy Feb 22 at 16:55

Assumptions: If the collision is elastic then: (1/2)m$${v_o}^2$$ + (1/2)I$${ω_o}^2$$ = (1/2)m$${v_f}^2$$ + (1/2)I$${ω_f}^2$$, and the friction is entirely static. The x axis is up the incline and y is up (but leaning from the vertical). $$ω_o$$ as a vector is toward -z (into the sketch). With static contact $$v_{fx}$$ = - R$$ω_f$$. (Where R is the radius of the ball and I is its rotational inertia.) The contact time is short enough to ignore the effect of gravity. The angular impulse (from friction) equals R times the linear impulse in the x direction: I(Δω) = Rm(Δ$$v_x$$). Expanding and solving the last equation gives: $$ω_f$$ = [I$${ω_o}$$ - mR$$v_{ox}$$]/[I – m$$R^2$$]. Knowing $$ω_f$$ lets you find $$v_{fx}$$ and putting both in the original energy equation yields $$v_{fy}$$.

• $v_{fx}$ = - R$ω_f$ is not a valid assumption for an elastic collision. This would mean that the contact point on the ball initially leaves the surface in a perpendicular direction. Any elastic collision must be time reversible--if the contact point can approach from a non-perpendicular direction, then it can leave along a non-perpendicular direction. – Ben51 Feb 22 at 17:12
• $v_{fx}$ = - R$ω_f$ applies to the component of the final velocity which is parallel to the surface, and does not rule out a perpendicular component. – R.W. Bird Feb 23 at 15:21
• If I understand your notation correctly, you are saying that the point of contact in the ball leaves the surface of the ramp in a perpendicular direction: the sum of the motions in the direction of the surface of the ramp due to rotation and translation are zero immediately after the collision. That’s not right – Ben51 Feb 23 at 16:00
• If the ball is to leave the collision with the same total kinetic energy that it had on approach, no energy can be lost to friction or the deformation of the surfaces. This means that the friction can be only static, and both surfaces must have returned to their original shape when the ball loses contact. At the instant before losing contact, the contact is static and the component of velocity parallel to the surface must be Rω. I'm thinking the deformation forces may transfer energy from perpendicular to parallel (or visa-versa). – R.W. Bird Feb 24 at 14:35
• Your first two sentences are correct. Your conclusion is not. Think of what happens during contact as the ball hooking into the wall with a stiff spring (this is an analogy for the elastic deformation during the collision). The spring stretches out and snaps back. It is not true that the instant after breaking contact the relative along-ramp velocity of the contact point is zero. Rather, it’s the same magnitude and opposite direction of what it was the instant before contact. – Ben51 Feb 24 at 14:47

You have three unknowns: two velocity components and the rotation rate. So you need three equations to determine them. Here are my recommendations:

1. Conservation of energy (there is translational and rotational, and the total can't change.)

2. Conservation of angular momentum about the contact point. (This one has not been mentioned in the other answers. There is angular momentum both before and after the collision due to both the CM motion and the rotation about the CM, and the total can't change because the contact force provides no torque about the contact point.)

3. Normal force does not depend on friction force. Or in other words, the component of the impulse in a direction perpendicular to the ramp surface will be the same regardless of the rotation rate. This seems reasonable, and it is the assumption underlying @John Wick's Step 2. It implies that component of the linear momentum of the ball in the direction perpendicular to the ramp changes sign from before to after the collision, but keeps the same magnitude. This seems reasonable, but it is more of an assumption than any kind of law--you could imagine friction mechanisms that would violate this assumption.

• During the collision, the normal and friction are not constant so there is a total net torque on the ball. So there’s no conservation in angular momentum. – Boy Feb 23 at 1:33
• Also conservation of energy does not really apply here since there is an immovable mass here. – Boy Feb 23 at 1:35
• Both those assertions are false. Angular momentum is conserved about the point of contact. and energy is conserved—that’s what “elastically” means – Ben51 Feb 23 at 1:36
• Note: The equation for conservation of angular momentum about the contact point reduces to the impulse equation in my answer. – R.W. Bird Feb 23 at 15:54