Question

A man throws a ball weighing \[500\,gm\] vertically upwards with a speed of \[10\,m{s^{ - 1}}\].
(a) What will be its initial momentum?
(b) What would be its momentum at the highest point of its flight?

Answer 1

Hint: The mass \[\left( m \right)\] and velocity \[\left( v \right)\] of an item are used to calculate linear momentum. It is more difficult to halt an object with more momentum. \[p = m \times v\] is the formula for linear momentum. Conservation of momentum refers to the fact that the overall amount of momentum never changes.

Complete step by step answer:
(a) As we know that;
Momentum $\left( p \right)$ = Mass \[\left( m \right)\] $ \times $ Velocity \[\left( v \right)\]
Here, mass $\left( m \right) = 500g$
We can convert gram \[\left( {gm} \right)\] to kilogram \[\left( {kg} \right)\]. A kilogram has \[1000\,gm\]
$m = \dfrac{{500}}{{1000}} \\
\Rightarrow m = 0.5kg \\ $
Also, velocity is given as $\left( v \right) = 10m{s^{ - 1}}$
Now, putting these value in the above formula we get:
Momentum $\left( p \right)$ $ = 0.5 \times 10 = 5\,kg.m{s^{ - 1}}$
So, its initial momentum will be $5\,kg.m{s^{ - 1}}$.

(b) Its momentum at the highest point of its flight will be zero. As at the maximum height the velocity of the ball is zero.

Additional Information: This above equation is simply Newton's second law applied to a system of particles. If the total external force operating on the system is zero, the system is said to be in equilibrium. When the overall force exerted on a particle's system is equal to zero, the system's total linear momentum is constant or conserved. This is the law of total linear momentum conservation for a system of particles.

Note: If there is friction, gravity, or a net force, momentum is not preserved (net force just means the total amount of force). What this means is that if you act on an object, it will modify its momentum. This should be self-evident, since you're changing the object's momentum by increasing or decreasing its velocity.