elastic collisions......physics help needed!!?

Question

a 2000 kg car whose brakes have failed coasts at 30 mph toward a 1000 kg car whose brakes are locked in "park" mode. If the coefficient of sliding friction between the 1000 kg car and the road is 0.7, determine how long after the first collision the second collision occurs.

Then find the distance between collisions.

Will there be a third collsion? proove or disprove.

Answer 1

This shouldn't be an elastic collision, but I will need two independent equations for the resultant speeds of the cars, so I will use conservation of momentum and conservation of energy.

1 mph = 0.44704 m/s
the 2000 kg car has momentum
30*0.44704*2000

26822.4
After the collision
26822.4 = 2000*v2+1000*v3
the 2000 kg car, assuming it is still coasting, will have no acceleration so v2(t)=v2, and distance, d2(t)=v2*t

the 1000 Kg car, on the other hand, will have
F=m*a=-m*g*.7
a=-.7g
so
v3(t)=v3-.7*g*t
and
d3(t)=v3*t-.35*g*t^2
when the second collision occurs
d2(2)=d3(t)
v2*t=V3*t-0.35*g*t^2
(v3-v2)/(.35*g)=t

using conservation of energy
.5*2000*30^2*0.44704^2=
.5*2000*v2^2+.5 *1000*v3^2
359.72=2*v2^2+v3^2
from above
26.8224 = 2*v2+v3
v3^2=719-107*v2+4*v2^2
plugging in
=719-359.72-107*v2+6*v2^2
0=359-107*v2+6*v2^2
v2=13.352, 4.48
v3=26.8224- 2*v2
v3=.1184, 17.8624
The only root that makes sense is the one where v3>v2, otherwise, if v3
so t=(17.86-4.48)/(.35*9.81)

3.9 seconds

the distances are equal (hence the collision)
I will prove later that this is incorrect:
d3(3.9)=17.8624*3.9-.35*9.81*3.9^2
14.4
d2(3.9)=4.48*3.9
14.4

The first step to determine the second collision is to compute
v2(3.9) this is v2, 4.48
and v3(3.9)=17.86-.7*9.81*3.9
Actually, The 1000 kg car stopped before the second collision
v3(t)=0=17.86-.7*9.81*t
t=2.6 seconds
the distance at 2.6 seconds is
d3(2.6)=17.8624*2.6-.35*9.81*2.6^2
23.23 meters
So the 1000 kg stopped and the 2000 kg car hit it going 4.48 m/s

So another collision occurs
4.48*2=2*v2+v3
v3^2=20.07-17.92*v2+4*v2^2
and
2*4.48^2=2*v2^2+v3^2
so
0=-20.07-17.92*v2+6*v2^2

v2=-0.868, 3.85

Let's see if the positive root produces a greater v3

v3=2*4.48-2*3.85
=1.26. No third collision occurs.

j

Answer 2

♠ here I’m to assume fully elastic collisions;
momentum before Mb=m1*v0; momentum after collision Ma=m1*u1+m2*u2;
and according to universal law of conservation of momentum Mb=Ma; or;
▬ hey big kid r u sure you don’t want to respond?
♪ m1*v0 = m1*u1+m2*u2, where m1=2*m2 =2000kg, m2=1000kg, v0=30 mph, u1 is speed of m1, u2 is speed of m2;