13. -7 points Tipler6 8.P.058. My Notes Ask Your Teacher A 3.8 kg block moving w

Question

13. -7 points Tipler6 8.P.058. My Notes Ask Your Teacher A 3.8 kg block moving with a velocity of +4.7 m/s makes an elastic collision with a stationary block of mass 1.6 kg. (a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision. m/s (for the 3.8 kg block) m/s (for the 1.6 kg block) (b) Check your answer by calculating the initial and final kinetic energies of each block. J (initially for the 3.8 kg block) (initially for the 1.6 kg block) 3 (finally for the 3.8 kg block) J (finally for the 1.6 kg block) Are the two total kinetic energies the same? Yes No eBook Submit Answer Save Progress Practice Another Version

Explanation / Answer

m1= 3.8 kg, m2= 1.6 kg, u1= 4.7 m/s , u2= 0

m1 u1 = m1 v1 + m2 v2

3.8 ( 4.7 ) = 3.8 v1 + 1.6 v2

u1 + v1 = u2 + v2

4.7 + v1 = 0 + v2

3.8 ( 4.7) = 3.8 ( v2- 4.7) + 1.6 v2

35.72 = 5.4 v2

v2= 6.615 m/s

v1= 1.915 m/s

Initial kE of mass 1= 0.5 ( 3.8) (4.7^2) = 41.971 J

Initial kE of mass 2= 0

Final KE of mass 1= 1/2 ( 3.8) (1.915^2) = 6.97 J

final kE of mass m2 = 1/2 (1.6) (6.615^2) = 35 J

yes

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