A 10 kilogram block initially hangs at rest at the end of two 10 meter strings of negligible mass as shown on the left diagram above. A 0.009 kilogram bullet, moving horizontally with a speed of 1600 meters per second, strikes the block and becomes embedded in it. After the collision, the bullet/ block combination swings upward, but does not rotate. Use g=10 m/s 2
a)Calculate the speed v of the bullet/ block combination just after the collision. (m/s)
b)Calculate the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/ block combination immediately after the collision.
c)Calculate the maximum vertical height above the initial rest position reached by the bullet/block combination. (m)
help me, AP physics on part C!!!!!
To solve this problem, we can use the principle of conservation of momentum and conservation of energy.
a) To calculate the speed of the bullet/block combination just after the collision (v), we can use the principle of conservation of momentum.
Momentum before the collision = Momentum after the collision
Momentum of the bullet before the collision (mv) = Momentum of the bullet/block combination after the collision ((m + M)v')
where m is the mass of the bullet, M is the mass of the block, and v' is the velocity of the bullet/block combination after the collision.
We know the mass of the bullet (m = 0.009 kg) and the initial speed of the bullet (u = 1600 m/s).
The momentum before the collision is mv = (0.009 kg)(1600 m/s) = 14.4 kg·m/s.
Since the bullet becomes embedded in the block, the mass of the bullet/block combination after the collision is (m + M) = (0.009 kg + 10 kg) = 10.009 kg.
Therefore, we can rewrite the equation as:
14.4 kg·m/s = (10.009 kg) v'
Solving for v', we find:
v' = (14.4 kg·m/s) / (10.009 kg) = 1.439 m/s
So, the speed of the bullet/block combination just after the collision is 1.439 m/s.
b) To calculate the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/block combination immediately after the collision, we can use the principle of conservation of energy.
The initial kinetic energy of the bullet is given by:
K.E. initial = (1/2) m u^2
The final kinetic energy of the bullet/block combination immediately after the collision is given by:
K.E. final = (1/2) (m + M) v'^2
Substituting the given values, we have:
K.E. initial = (1/2) (0.009 kg) (1600 m/s)^2 = 1152 J (rounded to the nearest whole number)
K.E. final = (1/2) (10.009 kg) (1.439 m/s)^2 = 10.238 J (rounded to the nearest whole number)
Therefore, the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/block combination immediately after the collision is:
K.E. initial / K.E. final = 1152 J / 10.238 J = 112.66 (rounded to two decimal places)
c) To calculate the maximum vertical height above the initial rest position reached by the bullet/block combination, we can use the principle of conservation of mechanical energy.
The initial mechanical energy (Ei) is equal to the final mechanical energy (Ef).
Ei = Ef
The initial mechanical energy is given by:
Ei = Potential energy at rest + Kinetic energy at rest
Since the initial position of the bullet/block combination is at rest, the initial kinetic energy is zero.
Ei = mgh + 0
The final mechanical energy is given by:
Ef = Potential energy at maximum height + Kinetic energy at maximum height
At maximum height, the final kinetic energy is zero.
Ef = 0 + mgh'
where h' is the maximum vertical height reached by the bullet/block combination.
Equating the initial and final mechanical energy:
Ei = Ef
mgh = 0 + mgh'
Canceling out the mass and rearranging the equation, we find:
h' = gh
Substituting the known values, we have:
h' = (10 m/s^2)(10 m) = 100 m
Therefore, the maximum vertical height above the initial rest position reached by the bullet/block combination is 100 meters.