3 of 9 A 15.0 kg block is attached to a very light horizontal spring of force co

Question

3 of 9 A 15.0 kg block is attached to a very light horizontal spring of force constant 425 N/m and is resting on a smooth horizontal table. (See the figure below 15.0 kg 3.00 kg 8.00 ms Tap image to zoom .) 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. Submit Request Answer

Explanation / Answer

let,

m1=15 kg

force constant, K=425 N/m

m2=3 kg

initial speed of m1 is, u2=8 m/sec

final speed of m2 is, v2=2m/sec

a)

by using law of consrvation of momentum,

m1*u1+m2*u2=m1*v1+m2*v2

15*0+3*8=15*v1-3*2

===> v1=2 m/sec

initial speed of m1 is v1=2 m/sec^2

now by using law of consrvation of energy,

1/2*m1*v1^2 = 1/2*K*x^2

1/2*15*2^2 = 1/2*425*x^2

===> x=0.376 m

maximum compression, x=0.376 m

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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