# In which direction does a wall exert the reaction force if it is hit diagonally?

I am confused about action-reaction forces and inertia of moving objects. As an example, take a cricket ball a bowler throws to the pitch:

Suppose the ball is about to drop in the pitch with velocity $v$. (the straight black line is the surface of the pitch)

Suppose it drops at $O$ - then it will apply a diagonal force $F_{OA}$ to the surface. Normally, I think that the surface will react with a force in the $\vec{OC}$ direction - then the ball should move in that direction, but it doesn't, it moves in the $\vec{OB}$ direction.

Why does this happen? Is it because of the ball's inertia?

• There is no such thing as a "reaction force" or an "action-reaction force," except in the trivial sense that all forces occur in pairs, so any force at all could be labeled in this way. Force A can be considered a "reaction" to force C, or C can be considered the "reaction" to A.
– user4552
Oct 10, 2014 at 16:15
• By the way, OA/OB/OC are not force vectors. The only force vector is its weight, and it is acting vertically downwards. OA/OB/OC are velocity vectors.
– t.c
Oct 10, 2014 at 16:39
• @DG_: It's usually not a good idea to rush to accept an answer on stackexchange. The answer you've accepted is completely wrong.
– user4552
Oct 10, 2014 at 19:12
• @BenCrowell Have you read my answer? I am arguing on the premise that OA/OB/OC are not force vectors, but velocity vectors. 1. The only force on a free projectile - that is one that is not propelled - is its weight, pointing downwards. 2. Friction is negligible. OA would have to be only very slightly tilted, if not normal to the ground. 3. A force in the OC direction does not mean motion in the OC direction - but a tendency to change its motion towards OC - but OP is asking why did the ball not move in the direction of OC (velocity vector), & not why did the ball not bounce at a lesser angle.
– t.c
Oct 10, 2014 at 20:09

By the way, OA/OB/OC are not force vectors. The only force vector that is acting on the ball before (and after) it hits the ground is its weight, and it is acting vertically downwards. OA/OB/OC are velocity vectors.

Try to split OA up into its vertical and horizontal components. When the ball hits the floor, reverse the vertical component due to impulse force that accelerates the ball upwards, normal to the surface. The impulse force is much larger in magnitude than its weight, therefore the ball is able to bounce upwards. Now sum up your original horizontal component and the reversed vertical component, you will find that it is equivalent to OB.

• +1. Again to the point. I just only want to know how can friction influence the components???
– user36790
Oct 10, 2014 at 11:03
• The effect of friction is negligible since the contact time is a fraction of a second.
– t.c
Oct 10, 2014 at 11:04
• @t.c: Actually I suspect that is not true. The friction force is always up to normal force times coefficient of friction as long as the bodies move relative to each other or there is a force that tries to move them. What happens is that the ball will start to spin which removes the relative motion and stops the friction before the ball stops moving in horizontal direction. Oct 10, 2014 at 12:27
• @JanHudec yes, you are right. I admit I tried to present a simplified version of the scenario to address the OP's question, since he was asking why doesn't the ball rebound in the direction of OC instead of OB. Under special conditions (if the ball is rotating or if the object is a box) there is a chance the object rebounds in the direction of OC.
– t.c
Oct 10, 2014 at 13:35
• This is wrong. Newton's 3rd law applies to all forces in the same way. The 3rd-law partner of A is C, not B.
– user4552
Oct 10, 2014 at 16:17

The force between two colliding bodies is not in the direction of motion. The normal and parallel components have to be treated separately.

The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.

There will also be parallel force caused by friction, but it has to be treated separately for two reasons:

• The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.
• The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.

So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.

If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.

• This is all true, but I don't think it addresses the question.
– user4552
Oct 10, 2014 at 16:18

Normally, I think that the surface will react with a force in the OC→ direction

Yes, if the ball's force on the surface is in direction A, then by Newton's third law, the surface's force on the wall is in direction C. This is what Newton's third law says. The third law applies to all forces in the same way.

However, as pointed out by Jan Hudec in a comment, the ball's force on the surface can't quite be in direction A if the motion is as given. The directions of the two forces should be more like this:

The x components of these forces are frictional. The y components are normal forces.

then the ball should move in that direction, but it doesn't, it moves in the OB→ direction.

Objects don't move in the direction of the force vector. Objects accelerate in the direction of the force vector. In your example, the leftward component of force C decelerates the ball's motion in the x direction, while the y component stops the y motion and then reaccelerates the ball in the positive y direction.

• "while the y component stops the y motion" sir I couldn't get it clearly!
– DG_
Oct 10, 2014 at 17:22
• While yes, objects accelerate in direction of the force vector, the OC→ direction exactly opposes the velocity vector and therefore would stop and then reverse the motion. Which does not happen. Therefore the force can't be in OC→ direction. Guess what, it isn't. -1. Oct 10, 2014 at 17:24
• The thing is that the floor does not have anything against the horizontal component of motion (ignoring friction), only against the vertical component. Therefore according to third law of motion only vertical force is created. Oct 10, 2014 at 17:26
• @JanHudec: Thanks for pointing that out. I'll correct my answer.
– user4552
Oct 10, 2014 at 18:59
• @JanHudec: The thing is that the floor does not have anything against the horizontal component of motion (ignoring friction), only against the vertical component. Therefore according to third law of motion only vertical force is created. No, this is incorrect. What you're claiming would be a violation of Newton's 3rd law. Therefore according to third law of motion only vertical force is created. This doesn't make sense. You seem to be claiming that if a frictional force existed here, it would violate Newton's 3rd law.
– user4552
Oct 10, 2014 at 19:07

As gravity force is their in Y direction so momentum is not conserved but in X direction as MC is valid so no change in the velocity. if take the components we can see the impulse is in positive Y direction only i.e. direction of force. so acceleration due to collision is in positive Y direction but gravity is their in negative Y direction. in X direction it will continue to move with same velocity. so net velocity vector will be along OB while force is their in only in positive Y direction. if fraction is present then it will give a rotational motion to the object.

• The OP is asking about the impulsive force when the ball collides with the surface. Gravity is negligible during this short time period. The force between the ball and the surface has both a y component (normal force) and an x component (friction).
– user4552
Oct 10, 2014 at 19:08
• @BenCrowell: What do you mean gravity is negligible? It doesn't seem very negligible to me. Oct 10, 2014 at 19:14
• if velocity is very high then for a very short interval of time we can neglect the gravity effect. Oct 10, 2014 at 19:17
• i didn't get you clearly @Ben Crowell Oct 10, 2014 at 19:18