How Does the Velocity of a Cart Change When Water Hits It?

  • Thread starter Thread starter sapz
  • Start date Start date
  • Tags Tags
    Cart Hose Water
AI Thread Summary
The discussion centers on calculating the velocity of a cart when water is sprayed at it, considering the effects of elastic collisions. The user attempts to derive the cart's velocity as a function of time, v(t), using momentum conservation principles. A key point raised is that the mass of water hitting the cart is not equal to the mass exiting the hose due to the cart's motion, similar to the Doppler effect. The correct approach involves adjusting the mass flow rate to account for the relative motion between the water stream and the cart. Ultimately, the user seeks clarification on how to accurately calculate the mass striking the moving cart.
sapz
Messages
33
Reaction score
1

Homework Statement


A friend asking:

A cart and a water hose:

A water hose is spraying water at rate m_dot kg/sec, and at a given velocity V, on a cart with a mass M.
The water hit the cart, and they are bounced back from it in an elastic collision (same velocity in relation to the cart, but opposite direction).
What is the velocity of the cart as a function of time, v(t)?

Homework Equations




The Attempt at a Solution


"My way of solution was this:

Lets have a coordinate system that moves with the same velocity with the cart at a certain time. Delta_m would signify the small amount of water. Then we write the momentum at time t, and t+dt:

P(t)=Delta_m*(V-v(t))
P(t+dt)=Delta_m*(v(t)-V)+M*dv


P(t)=P(t+dt), so:
dv/(V-v(t)) = (2*Delta_m/M)


Delta_m = m_dot*dt, so:

dv/(V-v(t))=2m_dot*dt/M --> ln(V-v(t))=-2m_dot*t/M + C, and then we can get v(t) by using exp.

However, the final answer I've seen is v(t) = (2m_dot*t/M)/(1+2m_dot*t/M),
Which is obviously not an end result of an exp/ln function.

What is wrong with my solution?
How do I arrive at the right solution?"
 

Attachments

  • watermomentum.png
    watermomentum.png
    2.5 KB · Views: 557
Physics news on Phys.org
Welcome to PF, sapz!
sapz said:
Delta_m = m_dot*dt

Think about this statement. Remember, you're in a reference frame that's moving relative to the hose.
 
TSny said:
Welcome to PF, sapz!


Think about this statement. Remember, you're in a reference frame that's moving relative to the hose.

Still no bells ringing. Could you explain your hint?
 
The mass of water that exits the hose in time ##dt## is ##\dot{m}\:dt##, but that's not the mass that hits the cart in time ##dt##.
 
Last edited:
How so?
 
If I'm standing on the ground throwing balls at a rate of 1 per second toward the cart, then the cart would not receive balls at a rate of 1 per second because the cart is moving away from me. It's like the Doppler effect. The frequency of reception is less than the frequency of the source.
 
Oh, I see. So how can Delta m be calculated?
 
For the stream of water, the rate at which mass strikes a wall is proportional to the speed of the stream relative to the wall. What would be the ratio of the rate at which mass strikes the moving cart to the rate at which mass would strike a stationary wall?
 
Back
Top