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20 A. A 35.09 bullet fravelling ot 325.0mis. bullet travelling ot 325.0mis skrifes dnd d be comes a 5449 embedded in block of uoodl that is sitting table top. g on a horizoa tal De ter mine the Speed of the Hock wood after wood afer the bo llet hos be come enhe ddeb' in it

20 A. A 35.09 bullet fravelling ot 325.0mis. bullet travelling ot 325.0mis skrifes dnd d be comes a 5449 embedded in block of uoodl that is sitting table top. g on a horizoa tal De ter mine the Speed of the Hock wood after wood afer the bo llet hos be come enhe ddeb' in it

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A 3509 be comes bullef trouelling of embeddedd in 325.0mis 20 strifes dnd I be comes a 5u4g block of uood that is siting table top. horizoatal on a O.) De tor mine the Speced of the bock f wood after has be come enbe ddeb' in it r the bo llet
Transcribed Image Text:A 3509 be comes bullef trouelling of embeddedd in 325.0mis 20 strifes dnd I be comes a 5u4g block of uood that is siting table top. horizoatal on a O.) De tor mine the Speced of the bock f wood after has be come enbe ddeb' in it r the bo llet
Expert Solution
Concept

The linear momentum remains conserved in collision.

Using conservation of linear momentum

Pinitial = Pfinal

Where Pinitial = Total initial linear momentum, Pfinal = Total final linear momentum

Answer

Mass of bullet is m1 = 35 g = 0.035 kg

Mass of block is m2 = 544 g = 0.544 kg

 

The initial velocity of the bullet is v1 = 325.0 m/s

The initial velocity of the block is v2 = 0 m/s (at rest)

 

Let V be the velocity of the block after the bullet embedded in the block. 

Using conservation of the linear momentum

Pinitial = Pfinal

m1 v1 + m2 v2 = (m1 + m2) V

0.035 × 325 +  0.544 × 0 = (0.035 + 0.544) × V

11.375 = 0.579 × V

V = 19.646  m/s

 

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