Block 1 of mass m 1 slides along a frictionless floor and into a one-dimensional

Question

Block 1 of mass m 1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mass m 2 = 6.0m 1 . Prior to the collision, the center of mass of the two-block system had a speed of 6.0 m/s. Afterward, what are the speeds of (a) the center of mass and (b) block 2?

Explanation / Answer

we have m2=6m1. velocity od center of mass: Xcm=(x1m1+x2m2)/(m1+m2), Vcm=(v1m1+v2m2)/(m1+m2). we have Vcm=6 and v2=0--> 6=v1m1/7m1--> v1=42 m/sec. then use conservation of momentum: m1v1=m1u1+m2u2 and conservation od energy: m1v1^2=m1u1^2+m2u2^2 solve two equations and you will get velocity of block 1 and block 2. then find velocity of center of mass Vcm=(u1m1+u2m2)/(m1+m2).

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---