Question

A 5.00-g bullet is fired into a 500-g block of wood suspended as a ballistic pendulum. the combined mass swings up to a height of 8.00 cm. What was the magnitude of the momentum of the combined mass immediately after the collision

Answer 1

`p = 0.633\,(kg\cdot m)/(s)`

Step-by-step explanation:

The combined speed after collision can be determined by means of the Principle of the Energy Conservation:

`K_(A) + U_(g,A) = K_(B) + U_(g,B)`

`K_(A) = K_(B) + U_(g,B) - U_(g,A)`

`(1)/(2)\cdot (m+M)\cdot v_(A)^(2) = (m+M)\cdot g\cdot h`

`v_(A) = √(2\cdot g \cdot h)`

`v_(A) = \sqrt{2\cdot (9.807\,(m)/(s^(2)) )\cdot(0.08\,m)}`

`v_(A) \approx 1.253\,(m)/(s)`

The magnitud of the momentum of the combined mass immediately after the collision:

`p = (0.005\,kg + 0.5\,kg)\cdot (1.253\,(m)/(s) )`

`p = 0.633\,(kg\cdot m)/(s)`