3) Mass 1 is moving at 2 m/s in the +x direction and it collides inelastically with mass 2 of 17 kg moving at 13 m/s in the + x direction. 120 Joules of KE is lost in this collision. What is the average force on mass 1 if the collision takes place in 0.2 seconds? Answer in Newtons.
here,
mass 1 is m1
initial speed of 1 , u1 = 2 m/s
mass of second block , m2 = 17 kg
initial speed of 2 , u2 = 13 m/s
let the final speed of combination be v
using conservation of momentum
m1 * u1 + m2 * u2 = (m1 + m2) * v
m1 * 2 + 17 * 13 = (m1 + 17) * v .......(1)
and
the kinetic energy lost , KE = KEi - KEf
KE = (0.5 * m1*u1^2 + 0.5 * m2 * u2^2) - 0.5 * (m1 + m2) * v^2
100 = 0.5 * ( m1 * 2^2 + 17 * 13^2) - 0.5 * (m1 + 17) * v^2 ....(2)
from (1) and (2)
m1 = 11.93 kg
v = 1.83 m/s
time taken , t = 0.2 s
the average force on mass 1 , F = m1 * ( v - u1) /t
F = 11.93 * ( 1.83 - 2)/0.2 N = -10.1 N
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