When William Tell shot the apple off his son's head, the arrow remained stuck in the apple, which means the collision between the arrow and apple was totally inelastic. Suppose that the velocity of the arrow was horizontal at 75 m/s before it hit, the mass of the arrow was 45 g, and the mass of the apple was 190 g. Suppose Tell's son was 1.4 m high.
A) Calculate the velocity of the apple and arrow directly after the collision.
B) Calculate how far behind the son the apple and arrow landed on the ground.
(A) Use conservation of momentum to calculate the horizontal velocity component Vx of the apple with embedded arrow after the apple is hit.
45*75 = (45 + 190) Vx
Use the height of the boy (H) to calculate how long the apple takes to hit the ground. Call that time T.
(1/2)gT^2 = H
Then multiply V and T for the answer to part (B).
drwls,
i think your way of solving the problem is incorrect. my online homework rejected your answers.
From your equations I concluded:
Vx=33.5 m/s, Vy=0
V=(Vx^2+Vy^2)^(1/2) = 33.5 m/s
X=v*(2h/g)^(1/2) = (2*1.4/9.81)^(1/2)
= 17.9 m
Unfortunately, these answers are wrong.
Your Vx is wrong, and is not given by the formula I wrote. Your T is correct.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision in an isolated system.
A) To calculate the velocity of the apple and arrow directly after the collision, we need to find the total momentum before the collision and set it equal to the total momentum after the collision.
The momentum before the collision can be calculated as follows:
Momentum of the arrow = mass of the arrow * velocity of the arrow
Momentum of the arrow = 0.045 kg * 75 m/s
The momentum of the apple is zero since it is initially at rest.
The total momentum before the collision is equal to the momentum of the arrow:
Total momentum before collision = 0.045 kg * 75 m/s
Let's call the velocity of the apple and arrow after the collision "vf". Since the collision is totally inelastic, the two objects stick together and move as one.
The total momentum after the collision is given by:
Total momentum after the collision = (0.045 kg + 0.19 kg) * vf
By setting the total momentum before the collision equal to the total momentum after the collision, we can solve for "vf".
0.045 kg * 75 m/s = (0.045 kg + 0.19 kg) * vf
Solving for "vf", we find:
vf = (0.045 kg * 75 m/s) / (0.045 kg + 0.19 kg)
B) To calculate how far behind the son the apple and arrow landed on the ground, we first need to find the time it took for the apple and arrow to hit the ground after the collision. We can use the equation of motion for vertical motion:
vf = vi + gt
Where "vf" is the final vertical velocity of the apple and arrow after the collision (which we found in part A), "vi" is the initial vertical velocity (zero since the apple and arrow were initially at rest vertically), "g" is the acceleration due to gravity (approximately 9.8 m/s^2), and "t" is the time taken.
Solving for "t", we find:
t = vf / g
Now that we know the time it took for the apple and arrow to hit the ground, we can find the distance they traveled horizontally. The distance is given by:
Distance = horizontal velocity * time
Since the horizontal velocity before the collision was 75 m/s and the vertical time is given by "t" (which we just calculated), we can find the distance by multiplying these values.
Distance = 75 m/s * t
Now you can substitute the value of "t" into the distance formula to find the final answer.