A 40 gram bullet is fired horizontally from a gun with a momentum of 2.8 (kg*m/s) and embeds itself into a 300 gram block of wood initially at rest on a wooden horizontal surface. After this collision the wood block slides 15 meters before falling off a table with a height of 2m. (neglect air resistance)
-Draw a diagram of this scenario and label the parts with the different physics concepts in play.
-Determine the horizontal distance travelled by the block after falling off the table.
-What is the final velocity of the block.
-At what angle does the block hit the ground? (relative to the horizontal)
-How much time is elapsed between the bullet hitting the block and the block striking the ground?
.04 kg bullet + 0.3 kg block = .34 kg final
initial momentum = momentum after collision
2.8 = .34 v
v = 8.24 m/s sliding along table
That is the horizontal velocity component until it hits the floor
How long to fall ?
h = (1/2) g t^2
2 = 4.9 t^2
t = .639 seconds falling
how far horizontal in .639 seconds
x = u t = 8.24 * .639 = 5.26 meters
vertical speed component
v = g t = 9.81 (.639) = 6.27 m/s
so velociy is 8.24 i - 6.27 j
speed =sqrt(8.24^2 + 6.27*2)
= 10.35 meters/second
tan angle below horizontal = 6.27/8.24
so angle = 37.3 degrees below horizontal
how long on table
15 meters/8.24
= 1.82 seconds on table
+ .639 in air
= 2.46 seconds
Unfortunately, I cannot draw a diagram here, but I can guide you through the process of solving the problem.
1. Draw a diagram: You can draw a diagram with a table, a bullet, a wooden block, and the ground.
2. Determine the horizontal distance traveled by the block: Since there is no horizontal force acting on the block after the collision, the horizontal distance traveled by the block is equal to the horizontal speed of the block multiplied by the time it takes to fall off the table.
To find the horizontal speed of the block, consider the conservation of momentum. Initially, the bullet has a momentum of 2.8 (kg*m/s), while the bullet and the block together have a final momentum of (0.04 kg + 0.3 kg) * v, where 'v' is the velocity of the bullet-block system after the collision. Set these two equal and solve for 'v'.
Next, you need to find the time it takes for the block to fall off the table. Use the formula for the time of fall: t = √(2h/g), where 'h' is the height of the table (2m) and 'g' is the acceleration due to gravity (9.8 m/s^2). Finally, multiply the horizontal speed of the block with the time of fall to get the horizontal distance traveled by the block.
3. Determine the final velocity of the block: To find the final velocity of the block, you can use the equation of motion: v^2 = u^2 + 2as, where 'v' is the final velocity (0 m/s since the block stops after falling), 'u' is the initial velocity (found from the bullet-block collision calculation), 'a' is the acceleration (due to gravity, -9.8 m/s^2), and 's' is the vertical distance (2m). Solve for 'u' and find the magnitude of the velocity vector.
4. Find the angle at which the block hits the ground: Use the tangent of the angle formula: tan(theta) = vertical distance / horizontal distance. In this case, the vertical distance is 2m, and the horizontal distance is calculated in step 2.
5. Calculate the time elapsed between the bullet hitting the block and the block striking the ground: Add the time of fall calculated in step 2 to the time it takes for the bullet to embed itself into the block. The latter time can be found by dividing the mass of the bullet (40g) by the momentum (2.8 kg*m/s) to get the velocity of the bullet, and then dividing the length of the bullet (which is not given) by its velocity to get the time.
By following these steps, you should be able to find the answers to all the given questions.