Answer:
u = 20.33 m/s
Step-by-step 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