Quick physics (motion) question:

  • Thread starter Thread starter Skorge
  • Start date Start date
  • Tags Tags
    Motion Physics
AI Thread Summary
A car traveling at 65 mph and a baseball thrown at 60 mph in the same direction would result in the baseball's speed relative to the ground being 125 mph. This scenario illustrates the addition of velocities under Galilean relativity. However, while the baseball can initially exceed the car's speed, it will quickly decelerate due to factors like gravity and air resistance. The discussion also clarifies that momentum, which depends on mass and velocity, differs between the car and the baseball, as the car is significantly more massive. Overall, understanding these concepts is crucial for grasping motion in physics.
Skorge
Messages
1
Reaction score
0
Hello,

I'm new to the forums. I just have a quick question. Anyway, I got a question/scenario:

Let's say there's a car traveling at 65 mph...if i was to throw a baseball at 60 mph outside in the same direction the car is going, what would be the speed of the baseball? Would it actually travel faster then the car moving at 65 mph? Just a general question. If you can direct me to some law's this problem poses please do so.

Thank you
 
Physics news on Phys.org
If you were able to throw a baseball at 60 mph with respect to the car, then its speed with respect to the earth would be 65+60 mph.

This is called addition of velocities under Galilean relativity (as opposed to Einsteinian relativity, which must be used when speeds are much higher).

Welcome to the forums, by the way!
 
Well, I'm still very new to the physics field so I can't give you the equations/laws involved but the short answer to your question is "yes"

Part of the reason is that the ball already has the same momentum (speed) as the car. Any additional momentum added to the ball will increase its speed so it will go faster than the car. Now if you threw the ball at 60MPH faster than the car, it will almost immediately slow down due to gravity, friction with the air, lack of mass to maintain velocity over distance, etc. The only way that ball would continue to go at 125MPH would be if you were in a weightless vacuum (aka space)

If anything I have said is wrong, somebody please correct me. I am also interested in any equations dealing with this.
 
I'm sorry sniper but I have some problems with your reply that I sould point out for the OP's sake.

sniper061 said:
the ball already has the same momentum (speed) as the car

I'm afraid the ball does not have the same momentum as the car. You imply that momentum and speed are the same, they are not. Momentum is the product of an objects mass and velocity; a car is much more massive that a ball, therefore if they are traveling at the same velocity they will have very different momentums.

You also say that

sniper061 said:
lack of mass to maintain velocity over distance

This also has eroneous implications. You imply here that a very massive object will not acclerate when acted upon by an unbalanced force. What would happen if you through a bowling ball out of the window? Would it keep going forever?
 
Last edited:
Hootenanny said:
I'm sorry sniper but I have some problems with your reply that I sould point out for the OP's sake.



I'm afraid the ball does not have the same momentum as the car. You imply that momentum and speed are the same, they are not. Momentum is the product of an objects mass and velocity; a car is much more massive that a ball, therefore if they are traveling at the same velocity they will have very different momentums.


Thank you, I knew I had some of the terminology wrong but I couldn't think of the right word at the moment.
 
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