A 15.0 kg block is attached to a very light horizontal spring of force constant

Question

A 15.0 kg block is attached to a very light horizontal spring of force constant 325 N/m and is resting on a smooth horizontal table. (See the figure below (Figure 1) .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

Part A

Find the maximum distance that the block will compress the spring after the collision.(Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

Explanation / Answer

first

Momentum will be conserved in the system

assume to the right is the positive direction and the spring is on the left side of the block.

m(u) + m(u) = m(v) + m(v)

m(u - v) = m(v - u)

u = 0

v = m(u - v) / m

v = 3.00(8.00 - (-2.00)) / 15.0

v = 2.00 m/s

change in kinetic energy will equal change in spring potential

PS = KE
0.5 * k * x² = 0.5 * m * v²

x² = m * v² / k

x = sqrt(m * v² / k)

x = sqrt(15 (2²) / 325)

x = 0.429 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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