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 575 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.)

Enter your answer using three significant figures.

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

A 15.0 kg block is attached to a very light horizontal spring of force constant 575 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.)

Enter your answer using three significant figures.

x =   m  

SubmitMy AnswersGive Up

Provide FeedbackContinue

Figure 1 of 1

5.0 kg 3.00 kg 8.00 mls

Explanation / Answer

Here,

Mass of block ,m1 = 15 Kg

Force constant ,k = 575 N/m

m2 = 3 Kg

u2 = 8 m/s

v2 = - 2m/s

Let the speed of 15 kg block after the collision is v

Using conservation of momentum

m2 * (8 - (-2)) = m1 * v

3 *(10) = 15 * v

v = 2 m/s

Now , for maximum compression

Using conservation of energy

0.5 * k * x^2 = 0.5 * m2 * v^2

575 * x^2 = 15 * 2^2

solving for x

x = 0.323 m

the maximum compression is 0.323 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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