Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Answer:
Incomplete questions
Let assume we are asked to find
Calculate the induced emf in the coil at any time, let say t=2
And induced current
Explanation:
Flux is given as
Φ=NAB
Where
N is number of turn, N=1
A=area=πr²
Since r=2cm=0.02
A=π(0.02)²=0.001257m²
B=magnetic field
B(t)=Bo•e−t/τ,
Where Bo=3T
τ=0.5s
B(t)=3e(−t/0.5)
B(t)=3exp(-2t)
Therefore
Φ=NAB
Φ=0.001257×3•exp(-2t)
Φ=0.00377exp(-2t)
Now,
Induce emf is given as
E= - dΦ/dt
E= - 0.00377×-2 exp(-2t)
E=0.00754exp(-2t)
At t=2
E=0.00754exp(-4)
E=0.000138V
E=0.138mV
b. Induce current
From ohms laws
V=iR
Given that R=0.6Ω
i=V/R
i=0.000138/0.6
i=0.00023A
i=0.23mA
Yes yes it will............
