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Question Number 130066 by Don08q last updated on 22/Jan/21

Calculate the energy loss when a 5 k g

object moving at 5 {m s}^{- 1} collides head

- on with and sticks to a stationary

3 k g object .

[A n s w e r : 3.75 J]

I n e e d h e l p . How do I arrive at this

answer ?

Commented by mr W last updated on 22/Jan/21

a n s w e r i s w r o n g!

Answered by ajfour last updated on 22/Jan/21

M u = (M + m) v

e n e r g y l o s s = \frac{1}{2} {M u}^{2} - \frac{1}{2} (M + m) {(\frac{M}{M + m})}^{2} u^{2}

= \frac{{M u}^{2}}{2} (1 - \frac{M}{M + m})

= \frac{{m M u}^{2}}{2 (M + m)} = \frac{(3 k g) (5 k g) {(5 {m s}^{- 1})}^{2}}{2 (5 k g + 3 k g)}

= \frac{3 \times 125}{16} J = \frac{375}{16} J

= 23 ∙ 437 J

Commented by Don08q last updated on 23/Jan/21

T h a n k y o u S i r

Answered by mr W last updated on 22/Jan/21

m_{1} = 5 k g

m_{2} = 3 k g

v_{1 i} = 5 m / s

v_{2 i} = 0 m / s

v_{1 f} = v_{2 f} = v_{f}

m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}

\Rightarrow m_{1} v_{1 i} = (m_{1} + m_{2}) v_{f}

\Rightarrow v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1 i} = \frac{5 \times 5}{5 + 3} = \frac{25}{8} m / s

l o s s o f e n e r g y :

Δ E = \frac{1}{2} m_{1} v_{1 i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}

= \frac{5 \times 5^{2}}{2} - \frac{(5 + 3) \times 25^{2}}{2 \times 8^{2}} = 23.4375 J

Commented by Don08q last updated on 22/Jan/21

T h a n k y o u S i r

Answered by Don08q last updated on 22/Jan/21

someone sent me this solution

K E = \frac{p^{2}}{2 m}

since momentum is conserved,

m_{1} u_{1} = (m_{1} + m_{2}) v = p

\Rightarrow p = 5 k g \times 5 {m s}^{- 1} = 25 {k g m s}^{- 1}

{K E}_{i} = \frac{p^{2}}{2 m_{1}}

Missing \left or extra \right

Energy loss = {K E}_{i} - {K E}_{f}

= \frac{p^{2}}{2} (\frac{1}{m_{1}} - \frac{1}{m_{1} + m_{2}})

= \frac{{(25)}^{2}}{2} (\frac{1}{5} - \frac{1}{5 + 3})

= 23.4375 J

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