Question

A ball of mass 2kg moves with a velocity 2m/s and strikes a wall . If the ball bounces back from the wall along the same path with same speed. Find the change in momentum

Answer 1

Answer:

The magnitude of the change in momentum in 8 kg m/s

Explanation:

Given :

A ball of mass 2 kg moves with a velocity 2 m/s and strikes a wall. The ball bounces back from the wall along the same path with same speed.

To find :

the change in momentum

Solution :

The linear momentum is defined as the product of mass and velocity.

⇒ p =  mv

Here, the ball bounces back from the wall along the same path with same speed.

so, initial velocity = final velocity

 u = v = 2 m/s

Change in momentum = final momentum - initial momentum

     = mv - (-mu)

     = mv + mu

     = mv + mv [ ∵ u = v ]

     = 2mv

     = 2(2)(2)

     = 8 kg m/s

Answer 2

Answer:

• 8 kg m/s

Explanation:

$\star\:{\underline{\underline{\tt{\purple{GIVEN}}}}}:-$

• A ball of mass $\bf{2 \:kg}$ moves with a velocity $\bf{2\: m/s}$ and strikes a wall.
• The ball bounces back from the wall along the same path with same speed.

$\star\:{\underline{\underline{\tt{\orange{TO\: FIND }}}}}:-$

• Change in momentum

$\star\:{\underline{\underline{\tt{\green{SOLUTION}}}}}:-$

As we know that ball bounces back from the wall along the same path with same speed. (Initial velocity = final velocity)

• $\sf\red{u = v = 2 m/s}$

$\underline{\boxed{\sf{Change\: in\: momentum = Momentum_{\:(final)} - Momentum_{\:(initial)}}}}$

$\qquad\quad\rightarrow\:\:\:\:\sf\orange{ mv - (-mu)}$

$\qquad\quad\rightarrow\:\:\:\:\sf\blue{ mv + mu}$

$\qquad\quad\rightarrow\:\:\:\:\sf\orange{ mv + mv \:[ \because u = v ]}$

$\qquad\quad\rightarrow\:\:\:\:\sf\blue{ 2mv}$

$\qquad\quad\rightarrow\:\:\:\:\sf\orange{ 2(2)(2)}$

$\qquad\quad\rightarrow\:\:\:\:{\boxed{\bf{\blue{ 8 \:kg\: m/s}}}}$