Question

A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.

(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.

Answer 1

a)
`V=1.067\: m/s`

b)
`v=434.65\: m/s`

Step-by-step explanation:

a)

Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.

`(1)/(2)MV^(2)=(1)/(2)k\Delta x^(2)`

Where:

• M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)
• V is the speed of the system
• k is the spring constant (6.17*10² N/m)
• Δx is the compression of the spring (0.0634 m)

Then, let's find the initial speed of the bullet-block system.

`V^(2)=(k\Delta x^(2))/(M)`

`V=\sqrt{(6.17*10^(2)*0.0634^(2))/(2.17935)}`

`V=1.067\: m/s`

b)

Using the conservation of momentum we can find the velocity of the bullet.

`mv=MV`

`v=(MV)/(m)`

`v=(2.17935*1.067)/(0.00535)`

`v=434.65\: m/s`

I hope it helps you!