Answer:
The maximum height risen by the bullet-baseball system after the collision is 81.76 m.
Explanation:
Given;
mass of the bullet, m₁ = 0.033 kg
mass of the baseball, m₂ = 0.15 kg
initial velocity of the bullet, u₁ = 222 m/s
initial velocity of the baseball, u₂ = 0
let the common final velocity of the system after collision = v
Apply the principle of conservation of linear momentum to determine the common final velocity.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.033 x 222 + 0.15 x 0 = v(0.033 + 0.15)
7.326 = v(0.183)
v = 7.326 / 0.183
v = 40.03 m/s
Let the height risen by the system after collision = h
Initial velocity of the system after collision = Vi = 40.03 m/s
At maximum height, the final velocity, Vf = 0
acceleration due to gravity for upward motion, g = -9.8 m/s²
[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]
Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.
