A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take
g=9.8ms−2

Question

A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the ceiling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take
g=9.8ms−2

dumbcow · Answer

i think this is where you set kinetic energy of bullet equal to potential energy of block at height of 30 cm

$$\frac{1}{2} m_1 v^2 = m_2 g h $$

anonymous · Answer

please explain more

dumbcow · Answer

im not a physics guy so not great at explaining this

but you would just plug in the masses and solve for velocity

miracrown · Answer

Not quite this.... @dumbcow
This problem is actually 3 problems in 1.
Actually 2 problems in 1.
Over counted!

The first part of this problem involves applying conservation of momentum during the collison of the bullet with the block...

anonymous · Answer

am not reallly good in physics, someone please help

miracrown · Answer

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