A 50 g bullet is fired into a 2 kg ballistic gel at rest on a frictionless surface. The bullet embeds itself in the gel and begins moving at 3.5 m/s. What was the initial velocity of the bullet?

The initial velocity of the bullet was 3.5m per sec because on a frictionless surface a bullet can't be stopped as there is no force opposing the bullet and as the bullet must be fired in certain velocity, we can say that the initial velocity of the bullet was 3.5mper sec.

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asuwafohjames asuwafohjames · Answer 2 · 2021-12-10T12:07:40+01:00

The initial speed of the bullet is 143.5 m/s

Applying the law of conservation of momentum,

mu+m'u' = V(m+m').................. Equation 1

Where m = mass of the bullet, m' = mass of ballistic, u = initial speed of the bullet, u' = initial speed of the ballistic, V = common speed after impact.

From the question,

Given: m = 50 g = 0.05 kg, m' = 2 kg, u' = 0 m/s ( at rest), V = 3.5 m/s

Substitute these values into equation 2 and solve for u

0.05u+2(0) = 3.5(2+0.05)

0.05u = 3.5(2.05)

0.05u = 7.175

u = 7.175/0.05

u = 143.5 m/s.

Hence, The initial speed of the bullet is 143.5 m/s

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