Answer: D
Reduced impact time will increase the impact force
Explanation: Collision occurs when two or more bodies collide and exert forces on each other within a short time.
If a body of mass M moving with a velocity V collide with another body, the kinetic energy of the body is equal to the work done by the body.
That is, K.E = 1/2mv^2 = F × s
Where workdone = Force × distance
Make F the subject of formula
Mv^2/2s = F
But V = distance s/time t
Substitute for V
Ms^2/2t^2s = F
Ms/2t^2 = F
From the equation above, we can deduce that F is inversely proportional to the square of time.
Therefore, the reduced impact time will increase the impact force
Most appropriate unit of measurement should be centimetres, cm.
Answer:
v=u+at is the first equation of motion. In this v=u+at equation, u is initial velocity.
The correct answer to this is (A. Units Only).
It shows that there is a velocity of 35, but the units are missing.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²