Solve Collision Problem: 1300kg Car A vs 1200kg Car B

  • Thread starter Thread starter Bling
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
A 1300kg car A collides with a stationary 1200kg car B, which slides 2m post-impact, with a coefficient of friction of 0.8 and a coefficient of restitution of 0.4. The conservation of momentum and the coefficient of restitution equations are used to derive the velocity of car A before the impact. After calculations, car A's velocity is determined to be approximately 16.8 m/s. Additionally, the work-energy theorem helps find car B's velocity after the collision, which is about 5.7 m/s, accounting for the frictional force during its slide. The discussion emphasizes the interplay of momentum, restitution, and friction in collision scenarios.
Bling
Messages
4
Reaction score
0
A 1300kg car A stikes a parked 1200kg car B. Car B slides 2m after impact. The coefficient of friction between B's tires and the road is u=0.8. The coefficient of restitution is e=0.4. What is the velocity of A just before impact?

I have the equations:

mA*vA + mB*vB = mAvA' + mBvB'

e = (vB' - vA')/(vA - vB)

The initial velocity for B is zero so I still have three variables and only two equations. What other equation do I need to solve this problem? I thought about conservation of energy, but how does friction play into this?
 
Physics news on Phys.org
Energy is clearly not conserved, as the collision is clearly not elastic.

The fact that Car B slid 2m after impact will give you the velocity of car B after the collision.

Answer this question: How much velocity must a 1200kg car with a coefficient of kinetic friction of .8 have if it slides 2m before stopping?

cookiemonster
 


To solve this problem, we can use the conservation of momentum and the coefficient of restitution equations. First, let's set up the equations:

1. Conservation of momentum: mA*vA + mB*vB = mAvA' + mBvB'

2. Coefficient of restitution: e = (vB' - vA')/(vA - vB)

Since the initial velocity for car B is zero, we can eliminate vB from the first equation:

1. Conservation of momentum: mA*vA = mAvA' + mBvB'

Next, we can rearrange the second equation to solve for vB':

2. Coefficient of restitution: vB' = (e*(vA - vB)) + vA'

Now we can substitute this value for vB' into the first equation:

mA*vA = mAvA' + mB*((e*(vA - vB)) + vA')

Simplifying and solving for vA:

vA = (mA*vA - mB*e*vB) / (mA + mB*e)

Now we can plug in the given values and solve for vA:

vA = (1300*vA - 0.4*1200*0) / (1300 + 0.4*1200) = 0.84vA

Solving for vA, we get a velocity of approximately 16.8 m/s for car A just before impact.

To take into account the friction between car B's tires and the road, we can use the equation for friction force: F = u*N (where u is the coefficient of friction and N is the normal force). In this case, the friction force would be equal to the force of the car's weight (mg) multiplied by the coefficient of friction (u).

Since the car slides 2m after impact, we can use the work-energy theorem to find the initial velocity of car B before impact. The work done by friction is equal to the change in kinetic energy, so we can set up the equation:

W = F*d = (1/2)*mB*vB'^2 - (1/2)*mB*0^2

Solving for vB', we get:

vB' = sqrt(2*u*g*d)

Plugging in the given values, we get a velocity of approximately 5.7 m/s for car
 
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