Answer:
1. B. Impulse = Force × Time
2. A. The momentum of each ball changes, and the total momentum stays the same
3. -55 kg·m/s
4. B. 3.5 kg
5. C. 6.3 m/s
Explanation:
1. The impulse is the momentum change of an object due to a force applied for a given period
2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;
The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same
3. By the principle of conservation of linear momentum, we have;
The sum of the momentum before the collision = The sum of the momentum after collision
Given that the objects move together after the collision, the total momentum is therefore;
Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s = -55kg·m/s
4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s
Where;
m₁ = The mass of the first object
m₂ = The mass of the second object
The total momentum of the system = 250 kg·m/s
From momentum = Mass × Velocity, we have;
Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg
The mass (m₁ + m₂) = 5.0 kg
Given that m₁ = 1.5 kg, we have;
m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg
The mass of the second object = 3.5 kg
5. The mass of the cue stick = 0.5 kg
The velocity of the cue stick = 2.5 m/s
The mass of the ball = 0.2 kg
The initial velocity of the ball = 0 m/s
Given that total initial momentum = Total final momentum, we have;
0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0
0.5 kg × 2.5 m/s = 0.2 kg × v
v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25 m/s ≈ 6.3 m/s
