Two cars A and B slide on an icy road as they attempt to stop at a traffic light
ID: 2058085 • Letter: T
Question
Two cars A and B slide on an icy road as they attempt to stop at a traffic light . Take Ma=876 kg, Mb=2600kg, and the friction coefficient between the locked wheels of either car and the road equal to 0.0858. Car A succeeds in coming to rest at the light , but car B cannot stop and rear-ends car A. After the collision the cars slide to rest 29.2m and 7.8m beyond of the impact point, respectively for car A and car B. Both drivers had their brakes throughout the incident . Find the speed at which car B hit car A. Hint : use the skid distance to find the speed of each car just after the impact and use the law of conservation of linear momentum. ( Note that the latter does not strictly apply because of the wheel-road friction forces present during the collision.please round the final answer to 5significant fig.
Explanation / Answer
Car A :
Vf^2 - Va^2 = 2as
Substituting values,
0 - Va^2 = 2(a)(29.2)
-Va^2 = 58.4a -----------------------------(1)
By definition,
f = coeff of friction * normal force
f = 0.0858(876)(9.8)
As f = ma
0.0858(876)(9.8) =876a
a = 0.84084 m/sec^2
Substituting a =0.84084 in Equation 1,
-Va^2 = 58.4(-0.84084)
Va^2 = 49.105056
Va = 7.00749998 m/sec.
Car B,
As coefficient of friction for either car is the same, then each car will be accelerating at the same magnitude.
Vf^2 - Vb^2 = 2(-0.84084 )(7.8)
0 - Vb^2 =13.117104
Vb = 3.621754271 m/sec.
Using conservation of momentum,
MaV1 + MbV2 = MaVa + MbVb
where
V1 = velocity of a car A just before impact = 0 (since car A was already stopped)
V2 = speed of car B just before impact
Substituting values,
876(0) + 2600V2 = 876(7.00749998) + 2600(3.621754271)
2600V2 = 15555.13109
V2 = 5.98274273
V2 = 5.98274 m/sec. (speed of car B just before impact)
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