ch 8 # 4 Blocks A (mass 4.50 kg ) and B (mass 13.00 kg ) move on a frictionless,

Question

ch 8 # 4

Blocks A (mass 4.50 kg ) and B (mass 13.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A.

A) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
Find the maximum energy. (in J)

B) Find the velocity of A. (in m/s)

C) Find the velocity of B. (in m/s)

D) Find the velocity of each block after they have moved apart.
Find the velocity of A. (in m/s)

E) Find the velocity of B. ( in m/s)

Explanation / Answer

A)

Vai = velocity of block A before collision = 5 m/s

Vbi = velocity of block B before collision = 0 m/s

V = common velocity

using conservation of momentum

ma Vai + mb Vbi = (ma + mb) V

4.50 (5) + 13 (0) = (4.50 + 13) V

V = 1.3 m/s

using conservation of energy

initial Total KE = final Total KE + Espring

(0.5) ma V2ai + (0.5) mb V2bi = (0.5) (ma + mb) V2 + Espring

(0.5) (4.50) (5)2 + (0.5) (13) (0)2 = (0.5) (4.50 + 13) (1.3)2 + Espring

Espring = 41.5 J

b)

V = 1.3 m/s

c)

V = 1.3 m/s

d)

using conservation of momentum

ma Vai + mb Vbi = ma Vaf + mb Vbf

4.50 (5) + 13 (0) = 4.50 Vaf + 13 Vbf

Vaf = (22.5 - 13 Vbf )/4.50                            eq-1

Using conservation of KE

ma V2ai + mb V2bi = ma V2af + mb V2bf

4.50 (5)2 + 13 (0)2 = 4.50 V2af + 13 V2bf

112.5 = 4.50 ((22.5 - 13 Vbf )/4.50)2 + 13 V2bf

Vbf = 2.6 m/s

Vaf = (22.5 - 13 Vbf )/4.50 = (22.5 - 13 (2.6) )/4.50 = - 2.51 m/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

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