We will apply the conservation of linear momentum to answer this question.
Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Note that v₁ and v₁' is the velocity of m₁ before and after the collision.
Let's choose m₁ and v₁ to represent the bullet's mass and velocity.
m₂ and v₂ represents the wood block's mass and velocity.
The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'
m₁v₁ + m₂v₂ = m₁v' + m₂v'
m₁v₁ + m₂v₂ = (m₁+m₂)v'
Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.
m₁v₁ = (m₁+m₂)v'
Here are the given values:
m₁ = 0.005kg
v₁ = 500m/s
m₂ = 5kg
Plug in the values and solve for v'
0.005×500 = (0.005+5)v'
v' = 0.4995m/s
v' ≅ 0.5m/s
If two object experiences the same displacement at the same time, the two objects must have the same average velocity.
The average velocity of an object is defined as the change in displacement per change in time of motion. This can be written as follows;

If two object experiences the same displacement at the same time, it means that the change in displacement with time is constant.

Thus, we can conclude that if two object experiences the same displacement at the same time, the two objects must have the same average velocity.
Learn more here: brainly.com/question/23856383
Answer:
10.8 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Acceleration (a) = 5 m/s/s
Distance travelled (s) = 291 m
Time (t) taken =?
We can calculate the time taken for the car to cover the distance as follow:
s = ut + ½at²
291 = 0 × t + ½ × 5 × t²
291 = 0 + 2.5 × t²
291 = 2.5 × t²
Divide both side by 2.5
t² = 291 / 2.5
t² = 116.4
Take the square root of both side
t = √116.4
t = 10.8 s
Thus, it will take the car 10.8 s to cover the distance.
Answer:
The normal force experienced by the car is approximately 8223.2 N
Explanation:
The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction
The mass of the vehicle that moves around the curve, m = 900 kg
The incline of the curve, θ = 20°
The speed with which the vehicle moves around the curve, v = 12.5 m/s
The radius of the curve, R = 50 meters
We have;

Where;
θ = The angle of inclination of the road = 20°
N = The normal force experienced by the car
m = The mass of the car = 900 kg
v = The velocity with which the car is moving = 12.5 m/s
R = The radius of the curve around which the vehicle moves = 50 m

The normal force experienced by the car = N ≈ 8223.2 N.