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A 0.086 kg bullet is fired at 266 m/s and collides with a 6.3 kg wooden block that is initially stationary. The bullet embeds into the wood block such that the collision is perfectly inelastic. Use conservation of momentum to find the speed of the wooden block (with bullet embedded) at the instant after the collision.

Respuesta :

Answer:

The value of speed of wooden block at the instant after the collision   [tex]V_f =[/tex] 262.41  [tex]\frac{m}{s}[/tex]

Explanation:

Given data

Mass of block = 6.3 kg

Mass of bullet = 0.086 kg

Velocity of bullet [tex]V_0[/tex] = 266 [tex]\frac{m}{s}[/tex]

Final velocity of whole system is

[tex]V_f = M_{b}\frac{ V_0}{M_b + (M_{block})}[/tex]

Put all the values in above formula we get

[tex]V_f = \frac{(6.3)(266)}{(6.3 + 0.086)}[/tex]

[tex]V_f =[/tex] 262.41  [tex]\frac{m}{s}[/tex]

This is the value of speed of wooden block at the instant after the collision.

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