A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a speed of 1.7 m/s. Calculate the speed of the bullet before striking the block. Answer in units of m/s

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

0.0283u₁  + 5.004 x 0   =  1.7(0.0283  +  5.004)

0.0283u₁   =   8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

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