1) we have
p1= m1*v1 = 8.70kg*21.0m/s = 182.7 kg-m/s
2) from the conservation laws,we have
m1*v1 = (m1 + m2)*V so V = m1*v1/(m1 + m2) = 8.70*21.0/(59.7)m/s = 3.06 m/s
3) Ki = 1/2*8.7*(21)2= 1918.35 J
Kf = 1/2(m1 + m2)*(3.06)2 = 279.50
So ΔK = 279.50- 1918.35= - 1638.85 J
Pb. (7.50) The x coordinate of center of mass is
The x-coordinate for mass 1.31 kg =0
The x-coordinate for mass 1.50 kg = 0.5 m
The x-coordinate for mass 1.10 kg = 0.75 m
Hence, we get
Pb. (7.26)
Let v1f be the final speed of the moving puck and
v2f be the final speed of the puck at rest.
We have
v1f= v1i(2*m1)/(m1+m2)
v1f= 5.74(0.9)/(1.35)
v1f= 3.83 m/s east
v2f = v1i (m1-m2)/(m1+m2)
v2f = 5.74*(-0.45)/(1.35)
v2f = 1.91 m/s west
A One-Dimensional Inelastic Collision Block 1, of mass n-8.70 ke. moves along a fiictioaless air track...
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