09, 2018 11:00 PM dow |Frt dth\' Ans e 010 (part 1 of 2) 10.0 points A(n) 66.4 g
ID: 2038505 • Letter: 0
Question
09, 2018 11:00 PM dow |Frt dth' Ans e 010 (part 1 of 2) 10.0 points A(n) 66.4 g object moving to the right at 22.9 cm/s overtakes and collides elastically with a(n) 26.1 g object moving in the same Earth direction at 14.1106 cm/s Find the velocity of the first object imme diately after the collision. Answer in units of cm/s that te eling golf b the c at 17 011 (part 2 of 2) 10.0 points Find the velocity of the second object imme- diately after the collision. ) Answer in units of em/s 012 10.0 points A car with mass 1122 kg, moving at 35 m/s, after strikes a(n) 2062 kg car at rest. if tbe two cars stick together, with what speed do they move? A riß with Answer in units of m/s. 013 10.0 points with a 3211 kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 36.9 north of east and at a speed of 4.85 m/s What was the speed of the 3211 kg car before the collision? Answer in units of m/s 9Explanation / Answer
10.
In a perfectly elastic collision
Using momentum conservation
Pi = Pf
m1V1i + m2V2i = m1V1f + m2*V2f
given that m1 = 66.4 & m2 = 26.1
V1i = 22.9 cm/sec & v2i = 14.1106
66.4*22.9 + 26.1*14.1106 = 66.4*V1f + 26.1*V2f
1888.84666 = 66.4*V1f + 26.1*V2f
Now other condition of the elastic collisions is that
V1f - V2f = V2i - V1i
V1f - V2f = 14.1106 - 22.9
V1f - V2f = -8.7894?
1888.84666 = 66.4*V1f + 26.1*V2f
Now Solving both equation
V1f = V2f - 8.7894
1888.84666 = 66.4*(V2f - 8.7894?) + 26.1*V2f
V2f*(66.4 + 26.1) = (1888.84666 + 66.4*8.7894)
V2f = (1888.84666 + 66.4*8.7894)/(66.4 + 26.1)
V2f = 26.7293 m/sec
V1f = V2f - 8.7894 = 26.7293 - 8.7894
V1f = 17.9399 cm/sec
Q11
V2f = 26.7293 cm/sec
Q12
In this type of collision:
using moment conservation
Pi = Pf
m1v1 + m2v2 = M*V
where M = m1 + m2
V = (m1v1 + m2v2)/(m1 + m2)
V = (1122*35 + 2062*0)/(1122 + 2062)
V = 12.33 m/sec
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