Three masses, 3m, and 6m can slide along a frictionless horizontal one-dimension

Question

Three masses, 3m, and 6m can slide along a frictionless horizontal one-dimensional track

As each mass collides with the next, they stick together, until all three are joined.

1) both of the above collisions are...

a) elastic

b) inelastic

2) During the first collision above (between m and 3m) which one or MORE of the following is/are true?
a) both masses receive impulses of equal magnitude and opposite direction
b) both masses experience horizontal forces of equal magnitude and opposite direction
c) both masses undergo accelerations of equal magnitude and opposite direction
d) 3m gains the same amount of speed that m loses

e) 3m gains the same amount of kinetic energy that m loses

3) Find the final speed of the three joined masses

4) Find the total kinetic energy lost form the initial to the final configuration

Three masses, 3m, and 6m can slide along a frictionless horizontal one-dimensional track As each mass collides with the next, they stick together, until all three are joined. 1) both of the above collisions are... a) elastic b) inelastic 2) During the first collision above (between m and 3m) which one or MORE of the following is/are true? a) both masses receive impulses of equal magnitude and opposite direction b) both masses experience horizontal forces of equal magnitude and opposite direction c) both masses undergo accelerations of equal magnitude and opposite direction d) 3m gains the same amount of speed that m loses e) 3m gains the same amount of kinetic energy that m loses 3) Find the final speed of the three joined masses 4) Find the total kinetic energy lost form the initial to the final configuration

Explanation / Answer

1. b) Inelastic

2. a) & b). Because in a system due to inertia internal forces may arise, as in this case, but they tend to produce ZERO resultant force.

3. Since no external unbalanced force acts on the entire system, linear momentum can be conserved. Therefore, Pf = Pi. That is, (m+3m+6m)*vf = m*v0. Therefore, vf = (v0/10).

4. Kinetic Energy lost = Final K.E. - Initial K.E.
   = 1/2*10m*(v0/10)2 - 1/2*m*v02
= -(1/2)*(9/10)*mv02 = -(9/20)*m*v02
Thus, the loss in kinetic energy is (9/20)*m*v02

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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