A car of mass 1689-kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is
2 answers:
Answer:
5.16 m/s
Explanation:
mass of car, m1 = 1689 kg
mass of truck, m2 = 2000 kg
Velocity of truck after collision, v2 = 17 km/h = 4.72 m/s
Let the initial velocity of car is u1.
initial velocity of truck, v1 = 0
velocity of car after collision, v1 = ?
Use conservation of momentum
m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2
1689 x u1 + 2000 x 0 = 1689 x v1 + 2000 x 4.72
1689 u1 = 1689 v1 + 9444.4 .... (1)
As the collision is elastic, so coefficient of restitution is 1.
Use the formula for the coefficient of restitution.
e = 1
v1 - 4.72 = 0 - u1
v1 = 4.72 - u1
Substitute the value of v1 in equation (1)
1689 u1 = 1689 (4.72 - u1) + 9444.44
1689 u1 = 7972.08 - 1689 u1 + 9444.44
3378 u1 = 17416.52
u1 = 5.16 m/s
Thus, the speed of car before collision is 5.16 m/s.
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Answer:
D = 271.54 m
Explanation:
given,
1. car accelerates at 4.6 m/s² for 6.2 s
2. constant speed for 2.1 s
3. slows down at 3.3 m/s²
distance travel for case 1
using equation of motion
d₁ = 88.41 m
case 2
constant speed for 2.1 s now, we have to find velocity
v = u + at
v = 0 + 4.6 x 6.2
v = 28.52 m/s
distance travel in case 2
d₂ = v x t
d₂ = 28.52 x 2.1 = 59.89 m
for case 3
distance travel by the car
v² = u² + 2 a s
final velocity if the car is zero
0² = 28.52² + 2 x (-3.3) x d₃
6.6 d₃ = 813.39
d₃ = 123.24 m
total distance travel by the car
D = d₁ + d₂ + d₃
D = 88.41 + 59.89 + 123.24
D = 271.54 m