Consider a frictionless track as shown in Figure. A block of mass m1 = 5.00 kg i

Question

Consider a frictionless track as shown in Figure. A block of mass m1 = 5.00 kg is released from A. It makes a head-on elastic collision at B with a block of mass m_2 = 10.0 kg that is initially at rest. When m_1 goes down to B. find the momentum of m_1 at B After collision, find the velocity of m_1. After collision, find the velocity of m_2. Calculate the maximum height to which m_1 rises up after the collision. If m_2 arrive at C after 1 sec after collision, what is the distance between B and C

Explanation / Answer

A.

apply the law of conservation of energy, we get

m1gh = 0.5m1v12

v1 = sqrt[2gh] = sqrt[2*9.8*5] = 9.9 m/s

the momentum of the mass m1 is,

p1 = m1v1 = 5*9.9 = 49.5 kg.m/s

--------------------------------------------------------------------------------

b.

after collision, the velocity of mass m1 is,

v1f = [m1-m2/m1+m2]v1+[2m2/m1+m2]v2i

= [5-10/5+10]*9.9 + 0

= -3.3 m/s

--------------------------------------------------------------------------

c.

after collision, the velocity of mass m2 is,

v2f = [2m1/m1+m2]v1i+ [m2-m1/m1+m2]v2i

= [2*5/5+10]*9.9

= 6.6 m/s

---------------------------------------------------------------------

d.

aply the law of conservation of energy, wwe get

0.5m1v2 = m1gh

h = 0.5v2/g

= 0.5*3.3*3.3 / 9.8

= 0.555 m

= 0.6 m

  

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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