
What is the momentum of the car weighing $1500\,kg$ when the speed increases from $36\,km\,h{r^{ - 1}}$ to $72\,km\,h{r^{ - 1}}$ uniformly?
Answer
500.4k+ views
Hint: 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 $\dfrac{{\Delta P}}{{\Delta t}} \propto F$ 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 $kg\,m\,{s^{ - 1}}$
$P = mv$
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, $m = 1500\,kg$ .
Initial velocity of the car, $u = 36\,km\,h{r^{ - 1}}$ .
Changing this to $m\,{s^{ - 1}}$ we get,
$u = 36\, \times \dfrac{{1000}}{{3600}}$
$ \Rightarrow u = 10\,m\,{s^{ - 1}}$
Initial momentum of the car is given by ${p_i} = mu$
Substituting the values, we get
${p_i} = 1500 \times 10$
$ \Rightarrow {p_i} = 15000\,kg\,m\,{s^{ - 1}}$
Final velocity of the car, $v = 72\,km\,h{r^{ - 1}}$ .
Changing this to $m\,{s^{ - 1}}$ we get,
$v = 72 \times \dfrac{{1000}}{{3600}}$
$ \Rightarrow v = 20\,m\,{s^{ - 1}}$
Final momentum of the car is given by ${p_f} = mv$
Substituting the values, we get
${p_f} = 1500 \times 20$
$ \Rightarrow {p_f} = 30000\,kg\,m\,{s^{ - 1}}$
The change is momentum is given by $\Delta p = {p_f} - {p_i}$
Substituting the values we get,
$\Delta p = 30000 - 15000$
$ \Rightarrow \Delta p = 15000$
So, the change in momentum is $\Delta p = 15000\,kg\,m\,{s^{ - 1}}$
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 $P = mv$ .
Momentum is described as the product of mass and velocity of the object. It is a vector quantity. Its SI unit is $kg\,m\,{s^{ - 1}}$
$P = mv$
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, $m = 1500\,kg$ .
Initial velocity of the car, $u = 36\,km\,h{r^{ - 1}}$ .
Changing this to $m\,{s^{ - 1}}$ we get,
$u = 36\, \times \dfrac{{1000}}{{3600}}$
$ \Rightarrow u = 10\,m\,{s^{ - 1}}$
Initial momentum of the car is given by ${p_i} = mu$
Substituting the values, we get
${p_i} = 1500 \times 10$
$ \Rightarrow {p_i} = 15000\,kg\,m\,{s^{ - 1}}$
Final velocity of the car, $v = 72\,km\,h{r^{ - 1}}$ .
Changing this to $m\,{s^{ - 1}}$ we get,
$v = 72 \times \dfrac{{1000}}{{3600}}$
$ \Rightarrow v = 20\,m\,{s^{ - 1}}$
Final momentum of the car is given by ${p_f} = mv$
Substituting the values, we get
${p_f} = 1500 \times 20$
$ \Rightarrow {p_f} = 30000\,kg\,m\,{s^{ - 1}}$
The change is momentum is given by $\Delta p = {p_f} - {p_i}$
Substituting the values we get,
$\Delta p = 30000 - 15000$
$ \Rightarrow \Delta p = 15000$
So, the change in momentum is $\Delta p = 15000\,kg\,m\,{s^{ - 1}}$
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 $P = mv$ .
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