A-block-of-wood-of-mass-0-6kg-is-balanced-on-top-of-vertical-port-2m-high-A-10g-bullet-is-fired-horizontally-into-the-block-and-the-embedded-bullet-land-at-a-4m-from-the-base-of-the-port-Find-the-init

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Question Number 29212 by NECx last updated on 05/Feb/18

A block of wood of mass 0.6kg is balanced on top of vertical port 2m high.A 10g bullet is fired horizontally into the block and the embedded bullet land at a 4m from the base of the port.Find the initial velocity of the bullet.

A b l o c k o f w o o d o f m a s s 0.6 k g i s

b a l a n c e d o n t o p o f v e r t i c a l p o r t

2 m h i g h . A 10 g b u l l e t i s f i r e d

h o r i z o n t a l l y i n t o t h e b l o c k a n d

t h e e m b e d d e d b u l l e t l a n d a t a 4 m

f r o m t h e b a s e o f t h e p o r t . F i n d

t h e i n i t i a l v e l o c i t y o f t h e b u l l e t .

Commented by NECx last updated on 06/Feb/18

p l e a s e h e l p

Answered by mrW2 last updated on 07/Feb/18

v_0 =initial velocity of bullet v_1 =velocity of wood with bullet t=(√((2h)/g)) L=v_1 t=v_1 (√((2h)/g)) ⇒v_1 =L(√(g/(2h))) m_B v_0 =(m_B +m_W )v_1 ⇒v_0 =((m_B +m_W )/m_B )×v_1 =((m_B +m_W )/m_B )×L(√(g/(2h))) =((10+600)/(10))×4(√((10)/(2×2))) =122(√(10)) ≈386 m/s

v_{0} = i n i t i a l v e l o c i t y o f b u l l e t

v_{1} = v e l o c i t y o f w o o d w i t h b u l l e t

t = \sqrt{\frac{2 h}{g}}

L = v_{1} t = v_{1} \sqrt{\frac{2 h}{g}}

\Rightarrow v_{1} = L \sqrt{\frac{g}{2 h}}

m_{B} v_{0} = (m_{B} + m_{W}) v_{1}

\Rightarrow v_{0} = \frac{m_{B} + m_{W}}{m_{B}} \times v_{1}

= \frac{m_{B} + m_{W}}{m_{B}} \times L \sqrt{\frac{g}{2 h}}

= \frac{10 + 600}{10} \times 4 \sqrt{\frac{10}{2 \times 2}}

= 122 \sqrt{10}

\approx 386 m / s

Commented by NECx last updated on 08/Feb/18

n o w i t s u n d e r s t o o d . G r a c i a s!

Commented by mrW2 last updated on 08/Feb/18

d e n a d a, s e \tilde{n o r}!

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