A 6.g bullet travelling horizontally strikes a 2.994-kg block of wood sitting at
ID: 1408663 • Letter: A
Question
A 6.g bullet travelling horizontally strikes a 2.994-kg block of wood sitting at the edge of a table The bullet is lodged into the wood. The table height is 1.2 m, and the block hit the floor 60 cm from the table. Full points will be awarded for the correct answer to the following question: How fast was the bullet traveling before it hit the block of wood? If the full credit cannot be awarded, then partial credit will be awarded for the correct answers to the following questions: What is the mass of the block of wood with the bullet stuck in it? What is vertical component of the block's velocity immediately after the bullet entered into it? How long does it take for the block to reach the floor once it is over the edge of the table? What is the horizontal component of the block's velocity as it is falling? What was the momentum of the block immediately after the bullet got logged into it? What was the momentum of the bullet before it entered the block? What was the velocity of the bullet before it entered the block?Explanation / Answer
time taken by block to fall from the table,
h = 0.5gt^2
1.2 = 4.9t^2
t = sqrt(12/49) = 0.495 sec
horizontal speed v = 0.60 /0.495 = 1.212 m/s
speed of bullet = 1.212*(2994+6)/6 = 606 m/s
From the above solution following answers are found
i) 2.994+0.006 = 3 kg
ii) 0
iii)0.495 sec
iv) 1.212 m/s
v) 1.212*(2.994+0.006) = 3.636 kgm/s
vi)3.636 kgm/s
vii) 606 m/s