Question

A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.​

Answer 1

Step-by-step-Explanation:

Given:A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s.

To find: Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Solution:

Change in momentum(∆p) =Final momentum-Initial momentum

Mass (m) =200gm=0.2 kg

Initial velocity (u) = 10 m/s

Final velocity (v) = 5 m/s

Change in momentum(∆p) =mv-mu

Change in momentum(∆p) =m(v-u)

Change in momentum(∆p) =0.2(-5-10)

Change in momentum(∆p) =0.2×(-15) = -3 Kg m/s

Final answer:

Change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick is -3 Kg m/s or -3 Ns.

Magnitude of change in momentum is 3 Ns.

(In case of magnitude,we must ignore the sign)

Note*: Sign of finial velocity should be taken negative because of opposite direction.As, velocity is a vector quantity.

Hope it helps you.

To learn more:

Force is given as F = 10² + 10t - 20. Find the change in momentum in the interval t=0 to t= 10sec ?

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https://brainly.in/question/44456555

Answer 2

Given Question :-

A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Formula Used :-

Momentum is defined as product of mass of the system and its velocity. i.e Momentum = mass × velocity.

$\large\underline{\sf{Solution-}}$

Given that,

Mass of the hockey ball, m = 200 g

We know,

$\boxed{ \bf{ \:1 \: kg \: = \: 1000 \: g}}$

Therefore,

Mass of the hockey ball, m = 0.2 kg

Velocity of hockey ball, u = 10 m/s

Now,

Initial momentum of the ball is

$\rm \:  =  \: mu$

$\rm \:  =  \: 0.2 \times 10$

$\rm \:  =  \: 2 \: kg \: m \: {s}^{ - 1}$

Now, after struck,

Mass of the hockey ball, m = 0.2 kg

Velocity of hockey ball, v = - 5 m/s

[ - ve sign because direction changes ]

Final momentum of the ball is

$\rm \:  =  \: mv$

$\rm \:  =  \: 0.2 \times ( - 5)$

$\rm \:  =  \: - \: 1 \: kg \: m \: {s}^{ - 1}$

Hence,

The change in momentum is

$\rm \:  =  \: mv - mu$

$\rm \:  =  \: - \: 0.1 - 0.2$

$\rm \:  =  \: - 0.3 \: kg \: m \: {s}^{ - 1}$

Thus,

• The change in momentum is - 0.3 kg m/s.