950 - kg toyota collides into the rear end of a 2200-kg cadillac stopped at a red light. the bumpers lock, the brakes are locked and the two car skid forward 4.8 meteres before stopping. the police officer, knowing the the coefficient of kinetic friction between tires and road is 0.4, calculates the speed of the toyota impact. what was the speed of the car?

u = 20.33 m/s

Explanation:

given,

mass of Toyota car = 950 Kg

mass of Cadillac = 2200 Kg

distance to stop = 4.8 m

coefficient of friction = 0.4

initial speed of the Toyota = ?

we know,

F = ma

and frictional foce

F = μ N = μ m g

where N is normal force

now equating both the equation

ma  = μ m g

 a = μ g

 a = 0.4 x 9.8

 a = 3.92 m/s²

using equation of motion

v² = u² + 2 a s

v² = 0² + 2 x 3.92 x 4.8

v = 6.13 m/s

above given velocity is the combined velocity of the Toyota and Cadillac

now, using conservation of momentum

m u = (M + m) v

950 x u = (2200+ 950) x 6.13

u = 20.33 m/s

The speed of Toyota before impact is equal to u = 20.33 m/s

the assestment will probley apele with cause and effect

the bal will not change its velocity.