“Inertia is the tendency of an object to remain at rest or continue moving at a constant speed in a straight line unless acted upon by an external force.”
An example of inertia is a passenger in a moving car feeling pushed back into their seat when the car suddenly stops, as the passenger continues to move forward due to their inertia.
The momentum of an object is equal to the mass multiplied by the velocity of the object
Momentum = mass x velocity
p = m x v
The SI unit of momentum is Ns (Newton second)
An example of momentum is a moving train, which has a large amount of momentum due to its mass and velocity, and is difficult to stop or change direction.
p = mv
The change in momentum is equal to the force multiplied by the time interval for which it was applied.
Consider a body of mass (m), moving with initial velocity (vi) A force (F) acts on the body to produce acceleration (a) therefore the final velocity after time (t) will become (vf).
If p = mv and m is constant, then the change in velocity changes the momentum of the body.
initial momentum = final momentum
pi = pf
mvi = mvf
Therefore change in momentum is represented as
Δp = pf − pi
pf − pi = mvf − mvi
pf − pi = m(vf − vi)
Divide both sides with t
\(\frac{p_f – p_i}{t} = m\frac{v_f – v_i}{t}\)
∵ \(a= \frac{v_f – v_i}{t}\)
∴ \(\frac{\Delta_p}{t} = ma\)
∵ \(F= ma\)
\(\frac{\Delta_p}{t} = F\)
\(F=\frac{\Delta_p}{t} \)
The above expression shows momentum in terms of force. A
A stone is dropped from the top of a tower 100 metres high. Find the time it takes for the stone to reach the ground and its velocity just before it hits the ground. (Assume air resistance is negligible.)
Solution: Step 1: Write the known quantities and point out quantities to be found.
vi = 0 m/s (initial velocity)vf = ? (final velocity) a = 9.8 m/s^2 (acceleration due to gravity) t = ? (time taken to reach the ground) h = 100 m (height of the tower)
Step 2: Write the formulas and rearrange if necessary. h = vi*t + (1/2)at^2 (to find the time taken to reach the ground) vf = vi + at (to find the velocity just before hitting the ground)
Step 3: Put the values in the formulas and calculate. Using h = vi*t + (1/2)at^2, we get: 100 m = 0 m/s * t + (1/2) * 9.8 m/s^2 * t^2 Solving for t, we get: t = sqrt(2 * h / a) = sqrt(2 * 100 / 9.8) ≈ 4.52 s
Using vf = vi + at, we get: vf = 0 m/s + 9.8 m/s^2 * 4.52 s vf ≈ 44.3 m/s
Therefore, the stone takes approximately 4.52 seconds to reach the ground and its velocity just before hitting the ground is approximately 44.3 m/s.”
Formula:
\(\Delta_p=Ft\)