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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Question
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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