A small block of mass 5 kg initially rests on a track at the bottom of a circular, vertical loop-the-loop, which has a radius of 0.8 m. The surface contact between the block and the loop is frictionless. A bullet of mass 0.7 kg strikes the block horizontally with initial speed 120 m/s and remains embedded in the block as the block and bullet circle the loop.
(a) Speed of block&bullet immediately after impact
(b) Kinetic Energy of block&bullet when they get midway to the top of the loop
(c) Minimum initial speed Vmin of bullet if block&bullet are to successfully execute a complete circuit of the loop
(a) CONSERVATION OF MOMENTUM
(b) CONSERVATION OF ENERGY
(c) NOT SURE
On c, you need to find velocity at the top what will give a centripetal force equal to mg, because if it is less than mg, the block will fall down.
So mg=mrw^2?
(and how do I find the w at the top of the loop?)
mg=mv^2/r
v^2=rg at top, so use that velocity at the top to figure the minimum vleocity at the bottom. (ke at top=keatbottom+mgh
Wouldn't it be KEbottom = KEtop without the mgh term since there's no height at the bottom?
Actually, I think it should be:
KE of bullet@bottom = KE of block&bullet@top + PE of block&bullet@top
Is that right?
Yes, it is right if you put the plus sign where you did.
To answer these questions, let's analyze the situation using principles of conservation of momentum and conservation of energy.
(a) Speed of block & bullet immediately after impact:
To find the speed immediately after the impact, we can use the principle of conservation of momentum, which states that the total momentum before and after the collision remains the same.
Before the collision:
The block is at rest, so its momentum is zero (since momentum = mass x velocity, and velocity is zero).
The momentum of the bullet before the collision is given by: momentum = mass x velocity = 0.7 kg x 120 m/s.
After the collision:
Since the block and bullet stick together after the collision, we can treat them as a single system.
Let's assume their velocity after the collision is V. The mass of the block and bullet system is 5 kg + 0.7 kg = 5.7 kg.
Using the principle of conservation of momentum, we can equate the momentum before the collision to the momentum after the collision:
0.7 kg x 120 m/s = 5.7 kg x V.
Solving for V, we find:
V = (0.7 kg x 120 m/s) / 5.7 kg.
Calculating this, we get:
V ≈ 14.737 m/s.
So, the speed of the block and bullet immediately after impact is approximately 14.737 m/s.
(b) Kinetic Energy of block & bullet when they get midway to the top of the loop:
To find the kinetic energy at the midway point, we can use the principle of conservation of energy, which states that the total energy before and after an event remains constant.
At the midway point of the loop, the potential energy of the block and bullet system will be maximum, while the kinetic energy will be minimum (almost zero).
The initial kinetic energy of the block and bullet system is given by:
Kinetic energy = (1/2) x mass x velocity^2.
Substituting the values, we get:
Initial kinetic energy = (1/2) x (5.7 kg) x (14.737 m/s)^2.
Calculating this, we find:
Initial kinetic energy ≈ 597.96 J.
So, the kinetic energy of the block and bullet system when they get midway to the top of the loop is approximately 597.96 J.
(c) Minimum initial speed Vmin of the bullet to successfully execute a complete circuit of the loop:
To find the minimum initial speed of the bullet, we need to consider the point where the block and bullet just manage to complete the loop without falling off. At this point, the net force acting on the system should provide the centripetal force required for circular motion.
The minimum initial speed occurs when the weight of the block and bullet system provides the necessary centripetal force.
The centripetal force at the top is given by:
Centripetal force = mass x acceleration = mass x (velocity^2 / radius).
Substituting the values, we have:
Centripetal force = (5.7 kg) x (Vmin^2 / 0.8 m).
The weight of the block and bullet system is given by:
Weight = mass x gravity = (5.7 kg) x 9.8 m/s^2.
To ensure that the block and bullet can complete the loop, the centripetal force at the top must be greater than or equal to the weight. Therefore, we have:
(5.7 kg) x (Vmin^2 / 0.8 m) ≥ (5.7 kg) x 9.8 m/s^2.
Simplifying the equation, we get:
Vmin^2 / 0.8 ≥ 9.8.
Multiplying both sides by 0.8, we find:
Vmin^2 ≥ 9.8 x 0.8.
Taking the square root of both sides, we obtain:
Vmin ≥ √(9.8 x 0.8).
Calculating this, we find:
Vmin ≥ 3.137 m/s.
Thus, the minimum initial speed (Vmin) of the bullet required for the block and bullet to successfully execute a complete circuit of the loop is approximately 3.137 m/s.