Question

If a plug of iron with dimensions 1 m by 1 m by 0.5 m breaks off a dam with a density of 7.8 × 10³ kg/m³, and water with a density of 1 × 10³ kg/m³ pushes it, what is the velocity of the iron plug upon collision?

a) 0.7 m/s
b) 1.4 m/s
c) 2.1 m/s
d) 2.8 m/s

Answer 1

To find the velocity of the iron plug upon collision, we use the principle of buoyancy and the equation of motion. The velocity is approximately 0.7 m/s.

Step-by-step explanation:

To find the velocity of the iron plug upon collision, we can use the principle of buoyancy. The buoyant force acting on the plug is equal to the weight of the water displaced by the plug. The weight of the water displaced can be calculated by multiplying its density by the volume of the plug. Since the plug breaks off the dam, it will fall freely under gravity until it reaches the water. We can use the equation of motion to find the velocity of the plug when it enters the water.

Let's calculate the weight of the water displaced:

Weight of water displaced = density of water × volume of plug

Volume of plug = 1 m × 1 m × 0.5 m = 0.5 m³

Weight of water displaced = (1 × 10³ kg/m³) × (0.5 m³) = 500 kg

Now, let's calculate the velocity of the plug upon collision:

Using the equation of motion, we have:

v² = u² + 2as

Here, u is the initial velocity (0 m/s), s is the distance traveled, and a is the acceleration due to gravity (-9.8 m/s²).

Since the plug is falling freely, the distance traveled is the depth of the water, which is given by:

s = h = 500 kg × 9.8 m/s² ÷ (7.8 × 10³ kg/m³ × 1 × 1 m)

s ≈ 0.063 m

Now, we can substitute the values into the equation of motion:

v² = 0 + 2 × (-9.8 m/s²) × 0.063 m

v ≈ 0.7 m/s

Therefore, the velocity of the iron plug upon collision is approximately 0.7 m/s.