A 6.5 Kg block is traveling in the negative x direction at 4.4 m/s,and a 3.5 Kg

Question

A 6.5 Kg block is traveling in the negative x direction at 4.4 m/s,and a 3.5 Kg block is traveling n the positive x direction at 7.6m/s.
1) find the velocity of each block after the collision in thecenter of mass reference frame (assume a head on collision)
2) fin the velocity of each block after the collision in theoriginal reference frame
1) find the velocity of each block after the collision in thecenter of mass reference frame (assume a head on collision)
2) fin the velocity of each block after the collision in theoriginal reference frame

Explanation / Answer

Velocity of each block with original refrence frame aftercollision is v1=(m1-m2)/(m1+m2)*u1+(2m1)/(m1+m2)*u2 For 6.5kg block v1=(6.5-3.5)/(6.5+3.5)*-4.4+(2*6.5)/(6.5+3.5)*7.6 m/s For 3.5kg block v1=(2m1)/(m1+m2)*u1+(m2-m1)/(m1+m2)*u2 The velocity of CM is VCM=(vx,vy)=(m1v1+m2v2)/(m1+m2),0 Velocity of each block with CM refrence frame aftercollision is v1x=VCM-v1 v2y=VCM-v2 Hence we get by it. v1=(2m1)/(m1+m2)*u1+(m2-m1)/(m1+m2)*u2 The velocity of CM is VCM=(vx,vy)=(m1v1+m2v2)/(m1+m2),0 Velocity of each block with CM refrence frame aftercollision is v1x=VCM-v1 v2y=VCM-v2 Hence we get by it.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---