A 5.0-g bullet traveling horizontally at an unknown speed hits and embeds itself in a 0.195-kg block resting on a frictionless table. The block slides into and compresses a 180-N/m spring a distance of 0.15m before stopping the block and bullet.
Determine the initial speed of the bullet.
To determine the initial speed of the bullet, we can make use of the principle of conservation of momentum and the concept of elastic collisions.
First, we need to determine the momentum of the bullet before the collision. The momentum (p) of an object is calculated as the product of its mass (m) and velocity (v):
p = m * v
Given that the bullet has a mass of 5.0 grams (0.005 kg), we can assume its momentum before the collision as p1 = (0.005 kg) * v1, where v1 is the initial velocity of the bullet.
After the collision, the bullet and the block move as a single system. We can consider this as the bullet and the block having a combined mass (m2) given by:
m2 = mass of bullet + mass of block = 0.005 kg + 0.195 kg = 0.200 kg
Let's assume that the final velocity of the bullet and the block together after the collision is v2.
According to the principle of conservation of momentum, the total momentum before the collision (p1) should be equal to the total momentum after the collision (p2):
p1 = p2
m1 * v1 = m2 * v2
(0.005 kg) * v1 = (0.200 kg) * v2 --(1)
Now, we need to determine the change in kinetic energy of the system during the collision. Since the collision is elastic, the kinetic energy is conserved.
The initial kinetic energy of the system is calculated as:
KE1 = (1/2) * m1 * (v1)^2
The final kinetic energy of the system is calculated as:
KE2 = (1/2) * m2 * (v2)^2
Let's assume the spring constant (k) of the spring as 180 N/m. The compression of the spring (s) is given as 0.15 m. The work done on the spring (W) is calculated as:
W = (1/2) * k * (s^2)
Since the spring is compressed to a stop at its maximum compression, this work done should be equal to the change in kinetic energy of the system:
W = KE1 - KE2
Substituting the equations for KE1, KE2, and W, we get:
(1/2) * k * (s^2) = (1/2) * m1 * (v1^2) - (1/2) * m2 * (v2^2)
(1/2) * (180 N/m) * (0.15 m)^2 = (1/2) * (0.005 kg) * (v1^2) - (1/2) * (0.200 kg) * (v2^2) --(2)
Now, we have two equations (Equations 1 and 2) with two unknowns (v1 and v2). We can solve these equations simultaneously to find the values of v1 and v2.
Solving these equations will provide the initial velocity of the bullet (v1).
Find Us=1/2k(deltax)^2
=1/2(180)(.15)^2
=2.025
This is spring potential energy. This must have been conserved from kinetic energy. Therefore Us=K=1/2mv^2
2.025=1/2(.195+.005)(v^2)
v=4.5 m/s
This is explains the time that the bullet is imbedded in the wood. To find the speed of the bullet before it is imbedded in the wood, use momentum.
p=mv
p(f)=4.5*(.195+.005)
p(f)=0.9
p(i)=p(f)
m*v=p(f)
.005*v=0.9
v=180 m/s