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Question: What is the momentum of the car weighing \(1500\,kg\) when the speed increases from \(36\,km\,h{r^{ ...

What is the momentum of the car weighing 1500 kg1500\,kg when the speed increases from 36 km hr−136\,km\,h{r^{ - 1}} to 72 km hr−172\,km\,h{r^{ - 1}} uniformly?

Explanation

Solution

Momentum is described as a consequence of the second law of motion. Newton’s second law of motion states that the rate of change of momentum is directly proportional to the applied force in the direction of force. The mathematical expression is given as ΔPΔt∝F\dfrac{{\Delta P}}{{\Delta t}} \propto F where P represents the momentum of the body at a given time t and F is the force applied.
Momentum is described as the product of mass and velocity of the object. It is a vector quantity. Its SI unit is kg m s−1kg\,m\,{s^{ - 1}}
P=mvP = mv

Complete step-by-step answer:
We know that the product of mass and velocity of the object is called its momentum.
Mass of the car, m=1500 kgm = 1500\,kg .
Initial velocity of the car, u=36 km hr−1u = 36\,km\,h{r^{ - 1}} .
Changing this to m s−1m\,{s^{ - 1}} we get,
u=36 ×10003600u = 36\, \times \dfrac{{1000}}{{3600}}
⇒u=10 m s−1\Rightarrow u = 10\,m\,{s^{ - 1}}
Initial momentum of the car is given by pi=mu{p_i} = mu
Substituting the values, we get
pi=1500×10{p_i} = 1500 \times 10
⇒pi=15000 kg m s−1\Rightarrow {p_i} = 15000\,kg\,m\,{s^{ - 1}}
Final velocity of the car, v=72 km hr−1v = 72\,km\,h{r^{ - 1}} .
Changing this to m s−1m\,{s^{ - 1}} we get,
v=72×10003600v = 72 \times \dfrac{{1000}}{{3600}}
⇒v=20 m s−1\Rightarrow v = 20\,m\,{s^{ - 1}}
Final momentum of the car is given by pf=mv{p_f} = mv
Substituting the values, we get
pf=1500×20{p_f} = 1500 \times 20
⇒pf=30000 kg m s−1\Rightarrow {p_f} = 30000\,kg\,m\,{s^{ - 1}}
The change is momentum is given by Δp=pf−pi\Delta p = {p_f} - {p_i}
Substituting the values we get,
Δp=30000−15000\Delta p = 30000 - 15000
⇒Δp=15000\Rightarrow \Delta p = 15000
So, the change in momentum is Δp=15000 kg m s−1\Delta p = 15000\,kg\,m\,{s^{ - 1}}

Note: The actual definition of the momentum is that the rate of change of momentum is directly proportional to the force. We describe momentum as the product of mass and velocity of the object only when the acceleration is constant. If acceleration is varying with respect to time or distance then momentum cannot be calculated using P=mvP = mv .