A 806 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 682 kg. The two cars lock up and slide together with a speed of 23.1 km/h. What was the speed of the first car just before the collision?

42.6km/h

Explanation:

Step one:

given data

mass of the first car M1= 806kg

the velocity of the first car V1=?

mass of the second car mass M2=682kg

velocity of the second car V2= 0km/h Note the car is parked

common velocity V=23.1km/hNote: the two cars have common velocity

since the collision is inelastic:

Step two:

Required:

The velocity of the moving car

We know that the expression for the conservation of linear momentum for inelastic collision is given as

M1V1+M2V2=V(M1+M2)

substitute

806*V1+682*0=23.1(806+682)

806V1+0=23.1(1488)

806V1=34372.8

V1=34372.8/806

V1=42.6km/h

The speed of the first car just before the collision is 42.6km/h

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