Collision of Spheres X & Y: Impulse Magnitude

  • Thread starter blackout85
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In summary, when a car accelerates, the force of friction between the tires and the road is what causes the car to move. This force is proportional to the normal force of the road on the tires and acts in the opposite direction of the car's motion. In a collision between two objects, the impulse of one object on the other is equal to the impulse of the second object on the first, according to Newton's third law. This is true regardless of the masses or velocities of the objects involved.
  • #1
blackout85
28
1
When you step on the accelerator to increase the speed of your car, the force that accelerates the car is, the force friction of the road on the tires, the force of your foot on the accelerator, the force of the engine on the drive shaft, or the normal force of the road on the tires.

When the car begins to accelerate, new forces come into play. The rear wheels exert a force against the ground in a horizontal direction; this makes the car start to accelerate. When the car is moving slowly, almost all of the force goes into accelerating the car. This is why I think the answer is the force friction of the road on the tires because the car is exerting a force in the horizontal direction against the ground.


Sphere X, of mass 2kg is moving to the right at 10m/s. Sphere Y, of mass 4 kg, is moving to the left at 10m/s. The two spheres collide head on. The magnitude of the impulse of X on Y is: twice the magnitude of impulse of Y on X, 1/2 the magnitude of Y on X, 1/4 the magnitude of impulse, four times the magnitude of impulse of Y on X, or the same magnitude of impulse of Y on X.


I take it that it would be the same magnitude of impulse of X on Y as Y on X--> due to P(before) =P(after)


Thank you :biggrin:
 
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  • #2
I agree with your answers, but your reasoning is not quite correct. In part 1, consider that the car might have front wheel drive. Or 4WD. And is it not the force of the ground on the car that accelerates it ,as a consequence of Newton III? Same for part 2, think Newton III.
 
  • #3
Assume the car wheels rest on a horizontal surface and rotate without slipping for simplifcation.

When a tire rotates on a surface (without slipping), the point of the tire that is in contact with the surface remains stationary (otherwise it would slip). The point wants to move one direction, but something prevents it from doing so? Some force must be present. Static friction enables the point to be stationary (otherwise the car would slip and we would have kinetic friction). The direction of the static friction points in the opposite direction of the tangential velocity of the point in contact with the road.

PhantomJay,

While it true there is a normal force acting on the car, the normal force itself will not cause the car to accelerate on a horizontal surface as it points perpendicular to the direction of motion; however, the frictional force (static) points along a line parallel to motion and contributes to a non-zero net force and thus a net acceleration. The frictional force is proportional to the magnitude of the normal force, it not not proportional to the normal force vector.

If the frictional force were proportional to the normal force vector, you could mathematically show that friction can't do work! A mistake I once assumed...

You could also argue that in an isolated system of the car and the earth, the car will not accelerate without any of the given options.
 
Last edited:
  • #4
physics.guru said:
The frictional force is proportional to the magnitude of the normal force, it not not proportional to the normal force vector.
I didn't mean to imply that the normal force of the ground on the car accelerates the car forward. I meant to say that the horizontal force of the ground on the car (that is, the friction force), accelerates it. Newton III in its finest hour.
 

Related to Collision of Spheres X & Y: Impulse Magnitude

1. What is the definition of impulse magnitude in the collision of spheres X & Y?

Impulse magnitude is a measure of the change in momentum that occurs when two objects collide. In the collision of spheres X & Y, it refers to the force exerted on the spheres during the collision, which results in a change in their velocities.

2. How is impulse magnitude calculated in the collision of spheres X & Y?

To calculate impulse magnitude in the collision of spheres X & Y, you need to know the mass and velocity of each sphere before and after the collision. The impulse magnitude is equal to the change in momentum, which can be calculated by multiplying the mass of the sphere by its change in velocity.

3. How does the mass of spheres X & Y affect the impulse magnitude in their collision?

The mass of spheres X & Y directly affects the impulse magnitude in their collision. A heavier sphere will experience a larger change in momentum and therefore a larger impulse magnitude than a lighter sphere. This is because the heavier sphere has more mass and therefore requires more force to change its velocity.

4. Does the velocity of spheres X & Y have an impact on the impulse magnitude in their collision?

Yes, the velocity of spheres X & Y does have an impact on the impulse magnitude in their collision. A higher velocity will result in a larger change in momentum and therefore a larger impulse magnitude. This is because the spheres are moving faster and therefore have more momentum to be changed by the collision.

5. What factors can influence the impulse magnitude in the collision of spheres X & Y?

The impulse magnitude in the collision of spheres X & Y can be influenced by several factors, including the masses and velocities of the spheres, the angle at which they collide, and the elasticity of the collision. Additionally, external factors such as friction and air resistance can also affect the impulse magnitude in the collision.

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