How long does it take for the block to fall to the ground? You are given the height of the table....
In the time it takes for the block to fall to the ground, the block has moved 2 meters. So, you can compute the final speed for this.
An 6.00 g bullet is fired into a 2.30 kg block initially at rest at the edge of a frictionless table of height 1.00 m. The bullet remains in the block, and after impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.
I know that the equation to solve for the initial velocity needs to be ((m_1+m_2)*v_f)/m_1 where m_1 is the mass of the bullet and m_2 is the mass of the block. What i don't know is how to get the final velocity in other terms so I can solve it.
2 answers
First you have to calculate the time it takes for the block to hit the ground.
y(t)=vit+(1/2)gt^2
y(t)=1+(1/2)(-9.80)t^2
Solve for t: t=0.45s
With distance and time you can calculate velocity: d/t= 2.00/0.45= 4.44m/s
Use this velocity in calculating the initial velocity of the bullet.
Multiply the mass of the block times your velocity that you just calculated...2.30kg x 4.44m/s=10.2 now divide this number by the mass of your bullet- be sure that you have like terms.
10.2/.006kg=1702 m/s is the initial velocity of your bullet
y(t)=vit+(1/2)gt^2
y(t)=1+(1/2)(-9.80)t^2
Solve for t: t=0.45s
With distance and time you can calculate velocity: d/t= 2.00/0.45= 4.44m/s
Use this velocity in calculating the initial velocity of the bullet.
Multiply the mass of the block times your velocity that you just calculated...2.30kg x 4.44m/s=10.2 now divide this number by the mass of your bullet- be sure that you have like terms.
10.2/.006kg=1702 m/s is the initial velocity of your bullet
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