Answer:
Incomplete question
Check attachment for the diagram of the question and the masses of the car are given in the diagram
Explanation:
Given that,
Car B is initially at rest
Ub = 0m/s
Car A is moving at
Ua = 30km/hr
Ua = 30×1000/3600 = 8.33m/s
This is an inelastic collision, after the collision the car move together at a speed of V
From the diagram
Mass of car A Ma = 1730kg
Mass of car B Mb = 935kg
Time taken during collison t = 0.1s
A. Common velocity V?
Applying conversation of momentum
Momentum before collision = momentum after collision
Momentum is given as p=mv
Now, momentum before collision
P(before) = Ma•Ua + Mb•Ub
P(before) = 1730 × 8.33 + Mb × 0
P(before) = 14,416.67
P(after) =(Ma+Mb)V
P(after) = (1730 +935)V
P(after) = 2665V
Then,
P(after) = P(before)
2665V = 14,416.67
V = 14,416.67/2665
V = 5.41m/s
To km/h
V = 5.41 ×1000./3600 = 19.47km/hr.
B. Average acceleration of each car
Car A
Acceleration is given as
a = ∆V/t
a = V-Ua/ t
a = (5.41 - 8.333) / 0.1
a = -2.924/0.1
a = -29.24m/s²
The negative sign show that car A is decelerating
Car B
a = (V - Ub) /t
a = (5.41 - 0) / 0.1
a = 5.41/0.1
a = 54.1 m/s²
This is showing that car B is accelerating, and it is reasonable because car B was initially at rest
C. Reaction of each car exerted on the other.
Using newton second second law of motion
Car A on B
F = ma
Ra = 1730 × 29.24
Ra = 50580N
Car B on A
F = ma
Rb = 935 × 54.1
Rb = 50583.5 N
Rb ≈ 50580N
As expected, the two reaction are suppose to be equal but due approximation along the way cause a slight change.
From newtons third law, for every action their is always equal and opposite reaction
So we expect reaction of car A to be equal to reaction of car B
