Question

4 billiard balls, each of mass 0.5kg, all are travelling in the same direction on a billiard table, with speeds of \[2m{s^{ - 1}},{\text{ }}4m{s^{ - 1}},{\text{ }}8m{s^{ - 1}}{\text{ }}and{\text{ }}10m{s^{ - 1}}.\] What is the linear momentum of this system?

Answer 1

Hint: For answering this question we will first define what momentum (linear momentum ) is. Later we will apply a formula that relates mass of each billiard ball with the velocity of each billiard ball to find the linear momentum of the whole system.

Formula Used:
\[P = m \times v\]
Where P = momentum of body,
V=Velocity with which the body is moving
M= mass of the moving body.

Complete step by step solution:
Let us see what momentum is defined as: it is the quantity of motion of a moving body, measured as a product of its mass and velocity.
A system's linear momentum is just the sum of its constituent parts' linear momentum. As a result, all we have to do now is calculate the momentum of each ball:
Ball 1:
Mass of ball 1= 0.5kg
Velocity of this ball =2m/s,
Hence momentum of ball one will be given by:
\[P = {m_1} \times {v_1}\]
\[ \Rightarrow P = 0.5 \times 2\]
\[ \Rightarrow P = 1kg\,m{s^{ - 1}}....(1)\]
Let this be equation 1.
Ball 2:
Mass of ball 2 = 0.5kg
Velocity of this ball = 4m/s,
Hence momentum of ball one will be given by:
\[P = {m_2} \times {v_2}\]
\[ \Rightarrow P = 0.5 \times 4\]
\[ \Rightarrow P = 2kgm{s^{ - 1}}\]
Let this be equation 2:
\[ \Rightarrow P = 2kgm{s^{ - 1}}.....(2)\]
Now, ball 3:
Mass of ball 3= 0.5kg
Velocity of this ball =8m/s,
Hence momentum of ball one will be given by:
\[P = {m_3} \times {v_3}\]
\[ \Rightarrow P = 0.5 \times 8\]
\[ \Rightarrow P = 4kgm{s^{ - 1}}\]
Let this be equation 3
\[ \Rightarrow P = 4kgm{s^{ - 1}}.......(3)\]
Now at last ball 4:
Mass of ball 4= 0.5kg
Velocity of this ball =10m/s,
Hence momentum of ball one will be given by:
\[P = {m_4} \times {v_4}\]
\[ \Rightarrow P = 0.5 \times 10\]
\[ \Rightarrow P = 5kgm{s^{ - 1}}\]
let this be equation 4
\[ \Rightarrow P = 5kgm{s^{ - 1}}.....(4)\]
Now we will add up equations 1,2,3 and 4 so that we get the net linear momentum,
\[ \Rightarrow P = {m_1} \times {v_1} + {m_2} \times {v_2} + {m_3} \times {v_3} + {m_4} \times {v_4}\]
On putting the values we get
\[ \Rightarrow P = 1 + 2 + 4 + 5kgm{s^{ - 1}}\]
\[ \Rightarrow P = 12kgm{s^{ - 1}}\]
Hence the net linear momentum of the system is \[P = 12kgm{s^{ - 1}}\].

Note:
We can directly apply p=mv only when we know that momentum of the system is conserved. In this case nothing has been mentioned about the billiard balls colliding elastically or in-elastically, it is given that they are just travelling in the same direction, hence we have applied p=mv. Do not apply this formula if momentum is not conserved.