KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
-1m/s
Explanation:
We can calculate the speed of block A after collision
According to collision theory:
MaVa+MbVb = MaVa+MbVb (after collision)
Substitute the given values
5(3)+10(0) = 5Va+10(2)
15+0 = 5Va + 20
5Va = 15-20
5Va = -5
Va = -5/5
Va = -1m/s
Hence the velocity of ball A after collision is -1m/s
Note that the velocity of block B is zero before collision since it is stationary
Fine, lets do a retry of this.
Δd = -0.9m
v₁ = 0
v₂ = ?
a = -9.8 m/s²
Δt = ?
We can use the following kinematic equation and solve for Δt.
Δd = v₁Δt + 0.5(a)(Δt)²
Δd = 0.5(a)(Δt)²
2Δd = a(Δt)²
√2Δd/a = Δt
√2(-0.9m)/(-9.8 m/s²) = Δt
0.<u>4</u>28571428574048 = Δt
Therefore, it takes 0.4 seconds for the glass to hit the ground, or 0.43s as you said (even though I don't believe it follows significant digit rules)
