Understanding the Collision Problem
This question asks us to find the speed of a cart after a boy jumps onto it. We are given the mass of the boy, his initial horizontal velocity, and the mass of the stationary cart. The cart has frictionless wheels, which is an important detail because it implies that there are no external horizontal forces acting on the boy-cart system during the jump. This lack of external force means that the principle of conservation of linear momentum can be applied.
A collision where two objects stick together after impact and move as a single unit is known as a perfectly inelastic collision. In such collisions, kinetic energy is not conserved, but linear momentum is always conserved in an isolated system.
Applying the Principle of Conservation of Linear Momentum
The principle of conservation of linear momentum states that if no external forces act on a system, the total linear momentum of the system remains constant. In this scenario, the system consists of the boy and the cart. Since the wheels are frictionless, we can consider the horizontal motion of the boy-cart system as isolated, meaning the total momentum in the horizontal direction before the boy jumps onto the cart is equal to the total momentum of the boy and cart combined after the jump.
Let's define the variables:
- \(m_b\) = mass of the boy = 52 kg
- \(v_{bi}\) = initial horizontal velocity of the boy = 2 m/s
- \(m_c\) = mass of the cart = 3 kg
- \(v_{ci}\) = initial horizontal velocity of the cart = 0 m/s (stationary)
- \(v_f\) = final velocity of the boy and cart together after the jump
The total initial momentum of the system is the sum of the individual momenta of the boy and the cart before the jump:
\(P_{initial} = m_b v_{bi} + m_c v_{ci}\)The total final momentum of the system is the momentum of the combined boy-cart system after the jump. Since they move together, their final velocity is the same:
\(P_{final} = (m_b + m_c) v_f\)According to the principle of conservation of linear momentum:
\(P_{initial} = P_{final}\) \(m_b v_{bi} + m_c v_{ci} = (m_b + m_c) v_f\)Step-by-Step Calculation of the Final Speed
Now, let's substitute the given values into the conservation of momentum equation:
\((52 \, \text{kg}) \times (2 \, \text{m/s}) + (3 \, \text{kg}) \times (0 \, \text{m/s}) = (52 \, \text{kg} + 3 \, \text{kg}) \times v_f\)Calculate the terms on the left side:
\(104 \, \text{kg} \cdot \text{m/s} + 0 \, \text{kg} \cdot \text{m/s} = 55 \, \text{kg} \times v_f\) \(104 \, \text{kg} \cdot \text{m/s} = 55 \, \text{kg} \times v_f\)Now, solve for the final velocity, \(v_f\):
\(v_f = \frac{104 \, \text{kg} \cdot \text{m/s}}{55 \, \text{kg}}\) \(v_f = \frac{104}{55} \, \text{m/s}\)To get a numerical value, perform the division:
\(v_f \approx 1.8909... \, \text{m/s}\)Rounding to two decimal places, the speed of the cart (and the boy) after the jump is approximately 1.89 m/s.
Comparing with Options
Let's look at the given options and compare them with our calculated speed:
| Option | Speed (m/s) | Comparison with 1.8909... m/s |
|---|---|---|
| 1 | 2.15 | Not close |
| 2 | 1.89 | Very close, matches the calculated value when rounded |
| 3 | 1.51 | Not close |
| 4 | 2.51 | Not close |
The calculated speed of approximately 1.89 m/s matches option 2.
Revision Table: Key Concepts Reviewed
| Concept | Description | Relevance to Problem |
|---|---|---|
| Linear Momentum | Product of mass and velocity (\(p = mv\)) | Used to quantify the motion of objects before and after the collision. |
| Conservation of Linear Momentum | Total momentum of an isolated system remains constant. | The core principle used to solve for the final velocity of the boy-cart system. |
| Isolated System | A system on which no external forces act. | Frictionless wheels ensure the horizontal system is approximately isolated, allowing momentum conservation. |
| Inelastic Collision | A collision where kinetic energy is not conserved, but momentum is (if isolated). Objects may stick together. | The boy jumping and landing on the cart is an example where they stick together. |
Additional Information: Extending Your Understanding
Let's explore a few related points to deepen your understanding of this physics problem and the concepts involved:
- Why is momentum conserved? Conservation of momentum is a direct consequence of Newton's third law of motion. During the interaction (the boy jumping onto the cart), the force exerted by the boy on the cart and the force exerted by the cart on the boy are equal in magnitude and opposite in direction. Over the small time interval of the interaction, the impulse (\(\text{Impulse} = F \Delta t\)) delivered by each object on the other is equal and opposite, leading to an equal and opposite change in momentum for each. The total change in momentum of the system is thus zero.
- What about energy? In this inelastic collision, kinetic energy is lost. This energy is typically converted into other forms, such as heat, sound, and deformation of the objects involved (though deformation might be minimal when jumping onto a rigid cart). You could calculate the initial kinetic energy (\(KE_{initial} = \frac{1}{2}m_b v_{bi}^2 + \frac{1}{2}m_c v_{ci}^2\)) and the final kinetic energy (\(KE_{final} = \frac{1}{2}(m_b + m_c) v_f^2\)) to see that \(KE_{final} < KE_{initial}\).
- What if there was friction? If there was significant friction in the wheels, the horizontal force of friction from the ground would be an external force acting on the system. In that case, linear momentum in the horizontal direction would not be conserved.
This problem is a classic example used to illustrate the power and application of the conservation of linear momentum in analyzing interactions like collisions and joins.
Important Questions from Conservation of Linear Momentum
- A particle of mass $6m$ at rest suddenly breaks on its own into three fragments.
Two fragments of mass $m$ and $2m$ move along mutually perpendicular directions with speeds $2v$ and $v$ respectively.
The energy released during the process is, - Body A of mass $m$ moving with speed $u$ collides with another body B of mass $3m$, at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is:
What are the dimensions of angular momentum?
The propulsion of a rocket is based on which of Newton's laws of motion?
For a system of interacting particles, which condition is fundamental for the conservation of its total linear momentum $\vec{P}$?
