A .035-kg bullet is fired vertically at 180 m/s into a .15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collicion, assuming the bullet embeds itself in the ball? Answer in meters.
initial momentum = .035*180 = 6.3 kg m/s
new mass = .035+.15 = .185 kg
new momentum = old momentum
6.3 = .185 v
v = 34.05 m/s
Ke = (1/2) m v^2 = (1/2)(.185)(34.05)^2
= 107.3 Joules
m g h = 107.3
h = 59.1
To find out how high the combined bullet and baseball rise after the collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if there are no external forces acting on it. In this case, the combined system of the bullet and baseball can be considered isolated because no external forces are acting on it during the collision.
First, let's calculate the initial momentum of the system before the collision.
The momentum of an object is given by the formula P = m * v, where P is momentum, m is mass, and v is velocity.
The initial momentum of the bullet is:
P_bullet_initial = m_bullet * v_bullet = 0.035 kg * 180 m/s = 6.3 kg*m/s
Since the baseball is initially at rest, its initial momentum is zero:
P_baseball_initial = m_baseball * v_baseball = 0 kg * 0 m/s = 0 kg*m/s
The total initial momentum of the system is the sum of the individual momenta:
P_initial = P_bullet_initial + P_baseball_initial = 6.3 kg*m/s + 0 kg*m/s = 6.3 kg*m/s
After the collision, the bullet embeds itself in the baseball. This means that the two objects stick together and move as one. Let's assume their combined mass after the collision is m_combined.
The final momentum of the combined system after the collision is:
P_final = m_combined * v_final
Since no external forces act on the system, the total momentum is conserved, meaning the initial momentum is equal to the final momentum:
P_initial = P_final
Substituting in the values, we get:
m_bullet * v_bullet = m_combined * v_final
Solving for v_final, we can rewrite the equation as:
v_final = (m_bullet * v_bullet) / m_combined
The final velocity of the combined system can be calculated using this equation.
To find the height the combined bullet and baseball rise, we can use the concept of conservation of mechanical energy. Neglecting air resistance, the gravitational potential energy before the collision is transformed into potential energy after the collision. The change in potential energy can be calculated using the equation:
ΔPE = m_combined * g * h
where ΔPE is the change in potential energy, m_combined is the combined mass of the bullet and baseball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Setting the change in potential energy equal to the initial kinetic energy of the system (which is the kinetic energy of the bullet before the collision), we have:
m_combined * g * h = 0.5 * m_bullet * v_bullet^2
Solving for h, we get:
h = (0.5 * m_bullet * v_bullet^2) / (m_combined * g)
Now we can substitute the given values to find the height:
m_bullet = 0.035 kg
v_bullet = 180 m/s
m_combined = m_bullet + m_baseball = 0.035 kg + 0.15 kg = 0.185 kg
g = 9.8 m/s^2
Plugging these values into the equation, we have:
h = (0.5 * 0.035 kg * (180 m/s)^2) / (0.185 kg * 9.8 m/s^2)
Calculating this expression will give you the height in meters.