A kitten of mass 0.60kg leaps 30° to the horizontal out of a toy truck of mass 1.2kg causing it to move over horizontal ground at 4.0m/s.At what speed did the kitten leap?

To solve this problem, we can apply the principle of conservation of momentum. This principle states that in a closed system, the total momentum before an event must equal the total momentum after the event, provided no external forces are acting on it. In this case, the system consists of the kitten and the toy truck.

Understanding the Components of Momentum

Momentum is calculated using the formula:

Momentum (p) = mass (m) × velocity (v)

Before the kitten leaps, both the kitten and the truck are stationary, so the initial momentum of the system is zero. After the kitten leaps, we need to consider the momentum of both the kitten and the truck.

Setting Up the Equation

Let’s denote:

• Mass of the kitten, m_k = 0.60 kg
• Mass of the truck, m_t = 1.2 kg
• Velocity of the truck after the leap, v_t = 4.0 m/s
• Velocity of the kitten, v_k (which we need to find)

Since the kitten leaps at an angle of 30°, we need to consider the horizontal component of its velocity. The horizontal component can be calculated using:

v_k,x = v_k × cos(30°)

Applying Conservation of Momentum

The total momentum before the leap is:

p_initial = 0

After the leap, the momentum of the kitten and the truck can be expressed as:

p_final = p_kitten + p_truck

Substituting the values, we have:

0 = (m_k × v_k,x) + (m_t × v_t)

Rearranging gives us:

m_k × v_k,x = - (m_t × v_t)

Calculating the Kitten's Velocity

Substituting the known values into the equation:

0.60 kg × (v_k × cos(30°)) = - (1.2 kg × 4.0 m/s)

Calculating the right side:

0.60 kg × (v_k × 0.866) = -4.8 kg·m/s

Now, solving for v_k:

v_k = -4.8 kg·m/s / (0.60 kg × 0.866)

v_k = -4.8 / 0.5196

v_k ≈ -9.23 m/s

The negative sign indicates that the direction of the kitten's leap is opposite to the direction of the truck's movement, which is expected since the kitten jumps out of the truck. Therefore, the speed at which the kitten leaped is approximately 9.23 m/s.

Final Thoughts

This problem illustrates the importance of understanding momentum and how it applies to real-world scenarios, such as a kitten leaping from a moving toy truck. By breaking down the components and applying the conservation of momentum, we can find the solution methodically.