87%-> Arabic-PC Sun 9:15 PM hapter 8, part b For this assignment, you submit ans

Question

87%-> Arabic-PC Sun 9:15 PM hapter 8, part b For this assignment, you submit answers by question parts. The number of submissions remaining for each question part orly changes f yo submit or dango the answer. Assignment Scoring Your last submission is used for your score. 45 polints Tipler6 8.PO58. My Not A 2.9 kg block moving with a velocity of +4.9 m/s makes an elastic collision with a stationary block of mass 2.0 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 2.9 kg block) m/s (for the 2.0 kg block) (b) Check your answer by calculating the initial and final kinetic energies of each block. J (initially for the 2.9 kg block) 3 Cinitially for the 2.0 kg block) J (finally for the 2.9 kg block) 3 (finally for the 2.0 kg block) Are the two total kinetic energles the same? No eBook Submit Answer Save Progress Pracice Another Version My Notes .2 points Tple 6 8 PO57 20

Explanation / Answer

ELASTIC COLLISION

m1 = 2.9 kg                               m2 = 2 kg

speeds before collision

v1i = 4.9 m/s                             v2i = 0 m/s

speeds after collision

v1f = ?                         v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ((2.9-2)*4.9 + (2*2*0))/(2.9+2)

v1f = 0.9 m/s ( for the 2.9 kg block )

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f = ((2-2.9)*0 + (2*2.9*4.9))/(2.9+2)

v2f = 5.8 m/s ( for the 2.0 kg block )

=======================================

part (b)

KE1i = (1/2)*m1*v1i^2 = (1/2)*2.9*4.9^2 = 34.8 J

KE2i = (1/2)*m2*v2i^2 = 0

KE1f = (1/2)*m2*v1f^2 = (1/2)*2.9*0.9^2 = 1.2 J

KE2f = (1/2)*m2*v2f^2 = (1/2)*2*5.8^2 = 33.6 J

YES

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