Two 500g blocks of wood are 2.0 m apart on a frictionless table. A 10g bullet is fired at 400m/s?

toward the blocks. It passes all the way through the first block, then embeds itself in the second block. Th e speed of the first block immediately after ward is 6.0m/s.

What is the speed of the second block after the bullet stop

Momentum is conserved. Remember, p = mv, momentum = mass times velocity, and the mass of the bulled is 10 g = 0.010 kg. And the mass of both blocks is 500 g = 0.500 kg. So the momentum of the bullet is (0.010 kg)(400 m/s) = 4.00 kg m/s. Given that the first block moves at a speed of 6.0 m/s after impact. So the momentum of the first block is (0.500 kg)(6.00 m/s) = 3.00 kg m/s. So, the momentum of the second block after the bullet hits is 4.00 kg m/s - 3.00 kg m/s = 1.00 kg m/s. And dividing by its mass gives its speed. But since the bullet embeds itself in the block, its mass is the mass of the bullet plus the mass of the block, so it has become 510 g = 0.510 kg.

So the speed of the second block after impact is (1.00 kg m/s)/(0.510 kg) = 1.96 m/s. Hope you got it

johanrusli johanrusli · Answer 2 · 2019-12-31T06:22:47-05:00

The speed of the second block is about 2.0 m/s

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Further explanation

Let's recall Impulse formula as follows:

\(\boxed {I = \Sigma F \times t}\)

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

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Given:

mass of block = M = 500 g

distance between block = x = 2.0 m

mass of bullet = m = 10 g

initial speed of bullet = u = 400 m/s

final speed of first block = v_M = 6.0 m/s

Asked:

speed of the second block = v' = ?

Solution:

This problem is about Conservation of Momentum.

Firstly, we will calculate the speed of the bullet as it passes all the way through the first block as follows:

\(mu = mv + Mv_M\)

\(10(400) = 10v + 500(6.0)\)

\(4000 = 10v + 3000\)

\(10v = 4000 - 3000\)

\(10 v = 1000\)

\(v = 1000 \div 10\)

\(\boxed {v = 100 \texttt{ m/s}}\)

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Next, we could find the magnitude of the speed of the second block as follows:

\(mv = ( m + M ) v'\)

\(v' = \frac{m}{m + M} v\)

\(v' = \frac{10}{10 + 500} \times 100\)

\(v' = \frac{10}{510} \times 100\)

\(v' = \frac{1}{51} \times 100\)

\(\boxed {v' \approx 2.0 \texttt{ m/s}}\)

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics