Given
Car 1
m1 = 1300 kg
v1 = 20 m/s
m2 = 900 kg
v2 = -15 m/s
(Negative sign shows that direction of car 2 is opposite to car 1)
Procedure
As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.
Answer:

Explanation:
The acceleration of an object is the rate of change of velocity of the object.
Mathematically, it is calculated as:

where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
Acceleration is a vector, so it is important to also take into account the direction of the velocity.
For the particle in this problem, we have:
u = +48 m/s is the initial velocity (positive direction)
v = -92 m/s is the final velocity (negative direction)
t = 4.5 s is the time interval
Therefore, the average acceleration is

Answer:
18.03 N
Explanation:
From the fiqure below,
Using parallelogram law of vector
R² = 15²+5²-2×5×15cos(180-60)
R² = 225+25-150cos120°
R² = 250-150(-0.5)
R² = 250+75
R² = 325
R = √325
R = 18.03 N
Hence the resultant force of the object is 18.03 N
Answer:
The horizontal component of displacement is d' = 1422.7 m
Explanation:
Given data,
The distance covered by the truck, d = 1430 m
The angle formed with the horizontal, Ф = 5.76°
The displacement is a vector quantity.
The horizontal component of displacement is given by,
d' = d cos Ф
= 1430 cos 5.76°
= 1422.7 m
Hence, the horizontal component of displacement is d' = 1422.7 m