The air-track carts in the figure (Figure 1) are sliding to the right at 1.0 m/s

Question

The air-track carts in the figure (Figure 1) are sliding to the right at 1.0 m/s. The spring between them has a spring constant of 150 N/m and is compressed 4.1 cm. The carts slide past a flame that burns through the string holding them together. What is the speed of 100-g cart? Express your answer to two significant figures and include the appropriate units. What is the direction of the motion of 100-g cart? What is the speed of 300-g cart? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

initial momentum = (100 + 300)g * 1m/s = 400 g·m/s
final momentum = 400 g·m/s = 100g * u + 300g * v
4 = u + 3v for u, v in m/s
u = 4 - 3v

initial energy E = U + KE = ½kx² + ½mv²
E = ½ * 150N/m * (0.042m)² + ½ * 0.4kg * (1m/s)² = 0.3323 J
final energy E = 0.3323 J = ½ * 0.1kg * u² + ½ * 0.3kg * v²
Dropping units for ease and substituting for u:
0.3323 = 0.05(4 - 3v)² + 0.15v² = 0.05(16 - 24v + 9v²) + 0.15v²
0.3323 = 0.8 - 1.2v + 0.45v² + 0.15v²
0 = 0.6v² - 1.2v + 0.4677
quadratic with roots at v = 0.53 m/s, 1.47 m/s

Clearly v > 1 m/s, so v = 1.47 m/s larger mass

u = 4 - 3*1.47 = -0.41 m/s smaller mass (backwards)

So

A)

The speed of 100g cart = 0.41 m/s in left direction

B)

To the left direction

C)

The speed of 300g cart = 1.47 m/s in right direction

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.