Propelling a car

Classical Mechanics Level 5

For some odd reason, you decide to throw baseballs at a car of mass M M , which is free to move frictionlessly on the ground. You throw the balls at the back of the car at speed u u such that the total mass of the car and its contents increases by n n kg/s (assume the rate is continuous, for simplicity).

If the car starts at rest, find its position at t = 10 t = 10 sec assuming that the back window is open, so that the balls collect inside the car.

Details and Assumptions

  • n = 1 n=1
  • M = 30 kg M=30 \text{ kg}
  • u = 5 ms − 1 u=5 \text{ ms} ^{ -1 }


The answer is 6.847.

Ritesh Yadav by Ritesh Yadav , 24, India

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1 solution

Let, λ = d m d t \lambda = \dfrac{\mathrm{d}m}{\mathrm{d}t}

So, Conserving Momentum, ( m + λ Δ t ) ( v + Δ v ) = m v + λ Δ t u ⇒ Δ v = λ Δ t ( u − v ) m + λ Δ t \displaystyle\begin{aligned}( m+\lambda\Delta t)(v+\Delta v) &= mv + \lambda\Delta t u\\ \Rightarrow \Delta v &= \dfrac{\lambda\Delta t(u-v)}{m+\lambda\Delta t}\end{aligned}

Differentiating w.r.t time, a = λ ( u − v ) m F = λ ( u − v ) \displaystyle\begin{aligned} a &= \dfrac{\lambda(u-v)}{m}\\F &= \lambda(u-v)\end{aligned}

We know, F = m d v d t \displaystyle F=m\dfrac{dv}{dt}

And, m = m 0 + λ t m = m_0 + \lambda t

Therefore, ( m 0 + λ t ) d v d t = λ ( u − v ) \displaystyle \left(m_0 + \lambda t\right)\dfrac{dv}{dt}=\lambda(u-v)

Integrating, ∫ 0 t λ d t m 0 + λ t = ∫ 0 v d v u − v ⇒ ln ⁡ ( m 0 + λ t m 0 ) = ln ⁡ ( u u − v ) ⇒ v = λ t u m 0 + λ t s = ∫ 0 t λ t u m 0 + λ t d t \displaystyle\begin{aligned}\int\limits_0^t \dfrac{\lambda dt}{m_0+\lambda t} &= \int\limits_0^v \dfrac{dv}{u-v} \\\Rightarrow \ln \left(\dfrac{m_0+\lambda t}{m_0}\right) &= \ln\left(\dfrac{u}{u-v}\right)\\ \Rightarrow v &= \dfrac{\lambda tu}{m_0+\lambda t}\\ s&= \int \limits_0^t \dfrac{\lambda tu}{m_0+\lambda t}dt\end{aligned}

Substituting, s = ∫ 0 10 5 t 30 + t d t = 50 − 150 ln ⁡ [ 4 3 ] ≈ 6.847689132232 86088411714910 09258852744735 43365335841524 00014719760606... \displaystyle\begin{aligned}s&=\int_0^{10} \dfrac{5t}{30+t}dt\\&=50-150\ln\left[\dfrac{4}{3}\right]\\ &\approx 6.847689132232\\&~~~~~86088411714910\\&~~~~~09258852744735\\&~~~~~43365335841524\\&~~~~~00014719760606...\end{aligned}

∴ s = 6.848 \therefore \boxed{s=6.848}

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