Collision of Spheres X & Y: Impulse Magnitude

  • Thread starter Thread starter blackout85
  • Start date Start date
AI Thread Summary
The discussion centers on the forces involved in a car's acceleration and the principles of impulse during a collision between two spheres. It emphasizes that the frictional force between the tires and the road is crucial for acceleration, as it provides the necessary horizontal force, while the normal force does not contribute to forward motion. In the collision scenario, the impulse exerted by Sphere X on Sphere Y is equal in magnitude to the impulse exerted by Sphere Y on Sphere X, in accordance with Newton's third law. The conversation also clarifies that static friction is essential for preventing slipping and enabling acceleration. Overall, the principles of Newton's laws are central to understanding both the car's motion and the collision dynamics.
blackout85
Messages
27
Reaction score
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:
 
Physics news on Phys.org
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.
 
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:
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top