A 14-cm diameter champagne bottle rests on its side on top of a frictionless horizontal table. Suddenly, the cork pops and the bottle slides backward, covering a distance of 22.0 cm in 0.44 s. If the mass of the bottle is 500 times the mass of the cork, find the distance from the original position that the cork will land on the table. Neglect air resistance and assume that the cork is very small compared to the bottle.
A) 8.5 cm
B) 3000 cm
C) 85 m
D) 60 m

Question

A 14-cm diameter champagne bottle rests on its side on top of a frictionless horizontal table. Suddenly, the cork pops and the bottle slides backward, covering a distance of  22.0 cm in  0.44 s. If the mass of the bottle is 500 times the mass of the cork, find the distance from the original position that the cork will land on the table. Neglect air resistance and assume that the cork is very small compared to the bottle.
A)  8.5 cm
B)  3000 cm
C)  85 m
D)  60 m

Quizplus · Accepted Answer

The conservation of momentum principle applies here, stating that the total momentum before the cork pops is equal to the total momentum after the cork pops. Before the cork pops, both the bottle and the cork are at rest, so the total initial momentum is 0. After the cork pops, the momentum of the bottle and the cork must still sum to 0, because there are no external horizontal forces acting on the system.Given that the mass of the bottle is 500 times the mass of the cork, let's denote the mass of the cork as $m$ and the mass of the bottle as $500m$. The velocity of the bottle can be calculated using the distance covered and the time taken, which is $v_b = \frac{0.22\,m}{0.44\,s} = 0.5\,m/s$.Using the conservation of momentum:$$0 = m_{cork}v_{cork} + m_{bottle}v_{bottle}$$$$0 = m \cdot v_{cork} + 500m \cdot (-0.5\,m/s)$$$$v_{cork} = 250\,m/s$$The cork's horizontal velocity is $250\,m/s$. To find the distance it travels, we use the same time duration because in the absence of air resistance and given that the horizontal table is frictionless, the only force acting on both objects after the pop is their mutual interaction at the moment of the pop. Thus, the time of travel for the cork until it lands should be the same as the bottle's, which is $0.44\,s$.Distance traveled by the cork = $v_{cork} \times time = 250\,m/s \times 0.44\,s = 110\,m$.However, none of the provided options match the calculated distance. Given the logic and calculations presented, there seems to be a mistake in the interpretation of the options or in the calculation. The correct approach based on conservation of momentum and the given data should lead to a calculation of the cork's distance, but the discrepancy suggests a reevaluation of the calculation or the assumptions made, such as the time of flight being equal for both cork and bottle, which might not directly apply if considering the cork's significantly higher velocity and potential differences in how air resistance might realistically affect it, despite the instruction to neglect air resistance. The correct approach, adhering strictly to the conservation of momentum and the provided data, indicates a significant oversight in matching the calculated outcome with the provided choices, suggesting a need for clarification or correction in the premise or the interpretation of the results.

A 14-Cm Diameter Champagne Bottle Rests on Its Side on Top