Two objects slide over a frictionless horizontal surface. The object A with a mass of 5000g is moves at a speed of 4.5m/s toward the object B (mB= 2500g) which is initially at rest. After the collision, both objects are directed to 30O on either side of the original line of motion of the object A.
a) What are the final speeds of the two objects?
To find the final speeds of the two objects after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
Momentum before the collision = Momentum after the collision
Momentum (m_A * v_A) + Momentum (m_B * v_B) = Momentum (m_A * v_Af) + Momentum (m_B * v_Bf)
Where:
m_A and m_B are the masses of objects A and B, respectively.
v_A and v_B are the initial velocities of objects A and B, respectively.
v_Af and v_Bf are the final velocities of objects A and B, respectively.
2. Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(0.5 * m_A * (v_A)^2) + (0.5 * m_B * (v_B)^2) = (0.5 * m_A * (v_Af)^2) + (0.5 * m_B * (v_Bf)^2)
Now, let's plug in the given values and solve for v_Af and v_Bf.
m_A = 5000g = 5 kg (Converting mass from grams to kilograms)
m_B = 2500g = 2.5 kg (Converting mass from grams to kilograms)
v_A = 4.5 m/s (Initial velocity of object A)
Using the conservation of momentum equation, we can solve for v_Bf:
m_A * v_A + m_B * v_B = m_A * v_Af + m_B * v_Bf
5 kg * 4.5 m/s + 2.5 kg * 0 m/s = 5 kg * v_Af + 2.5 kg * v_Bf
⇒ 22.5 kg·m/s = 5 kg * v_Af + 2.5 kg * v_Bf
Now, let's use the conservation of kinetic energy equation to solve for v_Af and substitute the value of v_Bf:
(0.5 * m_A * (v_A)^2) + (0.5 * m_B * (v_B)^2) = (0.5 * m_A * (v_Af)^2) + (0.5 * m_B * (v_Bf)^2)
(0.5 * 5 kg * (4.5 m/s)^2) + (0.5 * 2.5 kg * 0 m/s)^2 = (0.5 * 5 kg * (v_Af)^2) + (0.5 * 2.5 kg * v_Bf)^2
⇒ 0.5 * 5 kg * (4.5 m/s)^2 = 0.5 * 5 kg * (v_Af)^2 + 0.5 * 2.5 kg * (v_Bf)^2
⇒ 0.5 * 5 kg * (4.5 m/s)^2 = 0.5 * 5 kg * (v_Af)^2 + 0.5 * 2.5 kg * v_Bf^2
⇒ 45.0 kg·m^2/s^2 = 0.5 * 5 kg * (v_Af)^2 + 0.5 * 2.5 kg * v_Bf^2
Since we have two equations and two unknowns (v_Af and v_Bf), we can solve this system of equations simultaneously to find the values of v_Af and v_Bf.