Solving a Frictionless Cart Problem with Rain

[Gregie666]

hi.
can anyone push me in the right direction with the followin problem, please?

Homework Statement

a cart is moving on a frictionless surface at a speed of $$V_0$$
the mass of the cart is M.
it suddenly starts to rain at time t=0. the rain is dropping vertically at a rate of q gk per second.
the cart collects all the rain drops that hit it.
express the speed of the cart as a function of time passed since it started raining.

Homework Equations

$$F = {{dp} \over {dt}}$$

The Attempt at a Solution

$$& F = {{dp} \over {dt}} \Rightarrow \cr & 0 = {{dm} \over {dt}}v(t) + {{dv} \over {dt}}m(t) \Rightarrow \cr & 0 = qv(t) + {{dv} \over {dt}}(M + qt) \Rightarrow \cr & qv = - {{dv} \over {dt}}(M + qt) \Rightarrow \cr & qvdt = (M + qt)dv$$

so i get this equation and i don't know how to solve it for v...


**how do i add line breaks to the latex??**

 

[weatherhead]

just like any differential equation, split the variables, put in the initial conditions, and rearrange for the variable you want.

$$\int { \frac {q}{M+qt}} dt = \int {\frac {1}{v}}dv$$

Try it that way

 

[m2003]

Hello! This is an interesting problem. To solve it, we need to consider the forces acting on the cart. Initially, the only force acting on the cart is its own inertia, which is given by its mass (M) and velocity (V_0). However, once it starts raining, there is an additional force due to the raindrops hitting the cart. This force is given by the mass of the raindrops (q) multiplied by the velocity of the cart (v).

Using the equation F = ma, we can set up the following equation:

M(dv/dt) = -qv

This equation shows that as the mass of the cart increases due to collecting raindrops, its acceleration (dv/dt) decreases.

To solve for v, we can integrate both sides with respect to time:

∫dv = - q/M ∫dt

Integrating both sides will give us:

v = -qt/M + C

Where C is a constant of integration. To find this constant, we need to use the initial condition given in the problem. At t=0, the cart is moving at a velocity of V_0.

Therefore, v(0) = V_0

Substituting this into our equation, we get:

V_0 = -q(0)/M + C

C = V_0

So our final equation for the velocity of the cart as a function of time is:

v(t) = -qt/M + V_0

This shows that as time passes, the velocity of the cart decreases due to the increasing mass from collecting raindrops.

I hope this helps! Let me know if you have any further questions.