l Red Pocket F3:16 10 Problem 2 Two blocks of masses m2.00 kg and m4.00 kg are r

Question

l Red Pocket F3:16 10 Problem 2 Two blocks of masses m2.00 kg and m4.00 kg are released from rest at a height o-5.00 mon a frictioaless track as shown in the figure below. When they meet o the level portion of the track, they undergo a head-on collision. -5 5 x (m) Figure S Problem 2 (a) At the instance sbowa, what is the z-coordinate of the center of mass for two blocks? (3pt) (b) What are the eelocities of blocks just before their collision? Note that the blocks move in opposite directions just before the collision. (4pt) (c) If the collision is elastic, what are the velocities of blocks m and immediately after the collision? (Spt) (d) To what heights will the blocks rise after the collision? (4pt) (e) Now, if the collision in (a) is completely inelastic (the blocks stick togetber), what is the velocity of the blocks immediately after the collision? (4pt) Bonus: How mach kinetic energy is lost in the inelastic collision in (e) (3pt) Student

Explanation / Answer

(a)x cordinate of center of mass=(m*5-m*5)/(2m)=0

(b)velocity of both will be same but in opposite direction

Magnitude=sqrt(2*9.8*5)=9.9 m/s

For m1,u1=9.9 m/s

For m2,u2=-9.9m/s

(c)for elastic collision

Velocity of m1=v1=(u1*(m1-m2)+2*m2*u2)/(m1+m2)=-16.5m/s

v2=u2*(m2-m1)+2*m1*u1/(m1+m2)=3.3m/s

(d)height of m1=v1^2/(2*9.8)=13.9 m

Height of m2=v2^2/(2*9.8)=0.56m

(e)for inelastic collision

Conserving momentum

2*9.9-4*9.9=6*V

We got

V=-3.3m/s

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