Final answer:
The post-collision speed of the combined system of a 4.2 kg rock placed onto a 5.8 kg cart initially moving at 52 cm/s (converted to 0.52 m/s) is calculated using conservation of momentum, resulting in a final speed of approximately 0.303 m/s.
Step-by-step explanation:
The subject matter involves the post-collision speed of a rock gently placed onto a moving cart, which is a classic example of a physics problem pertaining to the conservation of momentum. As there is negligible external force acting on the system (rock plus cart), the total momentum before and after the rock is placed should be the same, assuming that placing the rock does not impart any significant external force to the system.
To calculate the post-collision speed, use the equation for conservation of momentum:
m1v1 + m2v2 = (m1 + m2)vfinal,
where m1 is the mass of the cart, m2 is the mass of the rock, v1 is the initial speed of the cart, v2 is the speed of the rock (which is 0 since it is gently placed), and vfinal is the final speed of the combined system.
Plugging in the given values:
5.8 kg × 0.52 m/s + 4.2 kg × 0 m/s = (5.8 kg + 4.2 kg) × vfinal
Now, solve for vfinal:
vfinal = (5.8 kg × 0.52 m/s) / (5.8 kg + 4.2 kg) ≈ 0.303 m/s
The final post-collision speed of the cart and rock system is approximately 0.303 m/s.