Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

A

15.0 kg

package is moving at a speed of

10.0 m s^{- 1}

, vertically upward along the

y

-axis, when it explodes into three fragments: a

2.00 kg

fragment is shot upward with an initial speed of

20.0 m s^{- 1}

and a

3.00 kg

fragment is shot in the positive direction of

x

-axis, horizontal with an initial speed of

5.00 m s^{- 1}

. Find

(a)

the speed of the third fragment right after the explosion and

(b)

the total kinetic energy provided by the explosion.

Accepted Answer

Given: Mass of package, $m = 15.00 kg$ , velocity $v = 10.0 m s^{- 1}$ $(y - axis)$

Mass of the $1^{st}$ fragment, $m_{1} = 2.00 kg$ , velocity $5.00 m s^{- 1}$ $(y - axis)$

Mass of $2^{nd}$ fragment, $m_{2} = 3.00 kg$ , velocity $v_{2} = 5.00 m s^{- 1}$ $(x - axis)$

Mass of $3^{rd}$ fragment, $m_{3} = m - (m_{1} + m_{2}) = 15.0 - (2.00 + 3.00) = 10.0 kg$

Velocity along $x$ , velocity along $x - axis = v_{3 x}$

By law of conservation of momentum, as momentum is a vector quantity, so,

Initial momentum along $x - axis$ = Final momentum along $x - axis$

and initial momentum along $y - axis$ = Final momentum along $y - axis$

Now, along $x - axis$ ,

$m \times 0 = m_{2} \times v_{2} + m_{3} \times v_{3 x}$

$0 = 3.00 \times 5.00 + 10.0 \times v_{3 x}$

$v_{3 x} = - 1.5 m s^{- 1}$

Along $y - axis$ ,

$m \times v = m_{1} \times v_{1} + m_{3} \times v_{3 y}$

$15.0 \times 10.0 = 2.00 \times 20.0 + 10.0 \times v_{3 y}$

$v_{3 y} = \frac{110}{10} = 11.0 m s^{- 1}$

By vector addition, velocity of $3^{rd}$ fragment is,

$v_{3} = \sqrt{{(v_{3 x})}^{2} + {(v_{3 y})}^{2}} = \sqrt{{(- 1.50)}^{2} + {(11.0)}^{2}}$

$v_{3} = 11.1 m s^{- 1}$

Total kinetic energy provided by explosion is equal to the change in kinetic energy,

$∆ K E = Final K E - Initial K E$

$= (\frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} + \frac{1}{2} m_{3} {v_{3}}^{2}) - \frac{1}{2} m v^{2}$

$= (\frac{1}{2} \times 2.00 \times 20 . 0^{2} + \frac{1}{2} \times 3.00 \times 5 . 00^{2} + \frac{1}{2} \times 10.0 \times 11 . 1^{2}) - \frac{1}{2} \times 15.0 \times 10 . 0^{2}$

$= 303.6 J$

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