A bullet of mass 45 g is shot at a speed of 220 m/s into a 5.0-kg bag hanging from a string from the ceiling. The bag absorbs the bullet and begins to swing. To what maximum vertical height will it rise? Hint: Energy is not conserved as the bullet enters the bag but is conserved after the bullet comes to rest in the bag and the bag is swinging upward.
Is this right?
mv + Mv= v(m+M)
(.045)(200) + (5)(0)= v(.045+5)
9= v(5.045)
v= 1.7839444m/s (this will be the velocity of both at the bottom of the arc)
Step 2, use the conservation of energy law:
(.5)mv2=mgh
(.5)(5.045)( 1.7839444)2=(5.045)(9.8)h
(.5)(5.045)(3.182457622)=(5.045)(9.8)h
8.027749351=49.441h
h=.1623702868
Do I have the right idea? thanks for your help.
In your first equation, there are two different velocity variables: before and after. You should not use the same symbol for both. You have the right idea applying conservation of momemtum, but you used the wrong bullet velocity. It is 220, not 200.
You are also correct applying conservation of energy to the upward swing
Well, you seem to have the right idea, but let me put it in my own Clowntastic way to double-check your work.
First, let's calculate the initial momentum of the bullet and the bag when the bullet enters the bag.
The momentum of the bullet is given by the equation mv, where m is the mass of the bullet and v is its velocity. So, the momentum of the bullet is (0.045 kg)(220 m/s).
Now, when the bullet is absorbed by the bag, the bag and the bullet move as one unit. So, the momentum of the bag and the bullet combined is equal to the momentum of the bullet before it entered the bag, which is just mbullet * vbullet.
Next, let's calculate the velocity of the combined bag and bullet system after the bullet is absorbed. We can use the equation mv + Mv = V(m + M), where m is the mass of the bullet, v is its velocity, M is the mass of the bag, and V is the final velocity. Rearranging the equation, we have V = (mbullet * vbullet) / (m + M).
Now, we move on to finding the maximum vertical height the bag will rise. We can use the conservation of energy principle here. Initially, the system has kinetic energy due to the bullet's motion. After the bullet is absorbed, the system has potential energy due to the height the bag will reach.
Utilizing the equation (1/2)mv^2 = mgh, where m is the mass of the system (mbullet + Mbag), v is the final velocity (which we calculated in the previous step), g is the acceleration due to gravity, and h is the maximum height, we can solve for h.
So, as you can see, you're on the right track with your calculations! Keep clowning around with the numbers, and I'm sure you'll find the correct answer.
Yes, you have the right idea. Let's go through the steps to solve the problem.
Step 1: Conservation of momentum
You correctly used the principle of conservation of momentum to find the velocity of the bag and bullet together after the bullet is absorbed by the bag.
Using the equation:
m_bullet * v_bullet + M_bag * v_bag = (m_bullet + M_bag) * v_final
where:
m_bullet = mass of the bullet = 45 g = 0.045 kg
v_bullet = initial velocity of the bullet = 220 m/s
M_bag = mass of the bag = 5.0 kg
v_bag = initial velocity of the bag (which is zero)
v_final = final velocity of both the bag and bullet after the bullet is absorbed
Substituting the values, we get:
(0.045 kg) * (220 m/s) + (5.0 kg) * (0 m/s) = (0.045 kg + 5.0 kg) * v_final
0.045 kg * 220 m/s = 5.045 kg * v_final
v_final = (0.045 kg * 220 m/s) / 5.045 kg
v_final ≈ 1.979 m/s
Step 2: Conservation of energy
You correctly used the conservation of mechanical energy to find the maximum height the bag will reach after absorbing the bullet.
Using the equation:
(1/2) * (m_bullet + M_bag) * v_final^2 = (m_bullet + M_bag) * g * h_max
where:
m_bullet = mass of the bullet = 0.045 kg
M_bag = mass of the bag = 5.0 kg
v_final = final velocity of both the bag and bullet after the bullet is absorbed ≈ 1.979 m/s
g = acceleration due to gravity = 9.8 m/s^2
h_max = maximum vertical height
Substituting the values, we get:
(1/2) * (0.045 kg + 5.0 kg) * (1.979 m/s)^2 = (0.045 kg + 5.0 kg) * 9.8 m/s^2 * h_max
(0.045 kg + 5.0 kg) * (1.979 m/s)^2 = (0.045 kg + 5.0 kg) * 9.8 m/s^2 * h_max
Simplifying the equation, we get:
8.356 kg * 3.916 m^2/s^2 = 49.441 kg m/s^2 * h_max
32.708896 kg m^2/s^2 = 49.441 kg m/s^2 * h_max
Dividing both sides by 49.441 kg m/s^2, we get:
h_max ≈ 0.661 m
Therefore, the maximum vertical height the bag will rise after absorbing the bullet is approximately 0.661 meters.
Yes, you have the right idea. You've correctly applied the principles of conservation of momentum and conservation of energy to solve the problem.
In the first step, you used conservation of momentum to calculate the velocity (v) of the bullet and bag after the collision. You added the momentum of the bullet (mv) to the momentum of the bag (Mv) and set it equal to the momentum of the system after the bullet comes to rest in the bag. This gives you the equation (.045)(200) + (5)(0) = v(.045 + 5), which simplifies to 9 = v(5.045). Then you solved for v, which gave you the velocity of the bullet and bag at the bottom of the arc.
In the second step, you used conservation of energy to find the maximum vertical height (h) that the bag will rise to. You equated the initial kinetic energy of the system (bullet and bag) to the potential energy at the maximum height. The equation is (.5)mv^2 = mgh. You substituted the known values and the previously calculated value of v, and solved for h. Your calculation of h = 0.1623702868 is correct.
Therefore, your solution is correct. Good job!