Using the following given values:
Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s
Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span>
</span>
Obtaining Va2 via law of conservation of momentum:
total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67
Total kinetic energy before collision:
KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J
</span>Total kinetic energy after collision:
KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:
Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
Not sure I understand can you explain more
As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m
