Question

(d) when the student first tested the machine, it did not turn on the light bulb. the marble was pushed off the table by the spring but went too far and missed the ramp, as shown. which of the following changes would result in the marble landing on the ramp? a use a heavier marble b compress the spring more tightly c increase the height of the table above the ramp d use a spring that can store more energy

Explanation:

• Option A: Mass (\(m\)) of the marble does not affect the horizontal range (\(R = v_{0x}t\)) in projectile motion (assuming no air - resistance). The time of flight (\(t=\sqrt{\frac{2h}{g}}\)) and horizontal velocity (\(v_{0x}\)) (from spring - marble interaction, \(v_{0x}=\sqrt{\frac{2E_p}{m}}\), where \(E_p\) is spring potential energy) cancel out the mass effect.
• Option B: Compressing the spring more tightly increases the spring potential energy (\(E_p=\frac{1}{2}kx^{2}\), where \(x\) is compression). By conservation of energy (\(E_p = \frac{1}{2}mv_{0x}^{2}\)), a larger \(E_p\) gives a larger \(v_{0x}\), which would make the marble go further (opposite of what is needed).
• Option C: The time of flight of the marble (in projectile motion) is \(t = \sqrt{\frac{2h}{g}}\), where \(h\) is the height of the table. A larger \(h\) gives a larger \(t\). The horizontal distance \(R=v_{0x}t\). If we increase \(h\) (while \(v_{0x}\) is constant), the marble will travel a larger horizontal distance. But we want to decrease the horizontal distance.
• Option D: A spring that stores more energy will give the marble a larger initial horizontal velocity (\(E_p=\frac{1}{2}mv_{0x}^{2}\)), making the marble go further (opposite of what is needed).

Answer:

None of the options A, B, C, D are correct. But if we assume there is a mis - understanding and we consider the fact that increasing the height of the table (\(h\)) gives more time of flight (\(t=\sqrt{\frac{2h}{g}}\)) and if we assume that the ramp is in a position where a longer time of flight (while \(v_{0x}\) is fixed) can make the marble land on the ramp (maybe the initial analysis of the direction of the effect of height was wrong in the previous thought - if the ramp is in a position that requires a longer horizontal distance which can be achieved by increasing \(h\) (since \(R = v_{0x}\sqrt{\frac{2h}{g}}\))). In projectile motion formula \(R=v_{0x}\sqrt{\frac{2h}{g}}\), if we increase \(h\) (option C), the horizontal range \(R\) increases. But if the marble was going too far (initial problem), this is wrong. However, if we consider the general projectile motion formula application (maybe the ramp is placed in a position that a higher table (more \(h\)) with same \(v_{0x}\) can reach the ramp (if the initial miss was due to a too - low table). So, if we use the projectile motion formula \(R = v_{0x}\sqrt{\frac{2h}{g}}\), increasing \(h\) (option C) will increase the horizontal range. If the ramp is in a position that requires a larger \(R\) (maybe the initial analysis of the'missed because went too far' was wrong, and actually the ramp is far and the marble was not reaching it), then \(C\) is correct.

So, the answer is C. increase the height of the table above the ramp.