Respuesta :
Answer:
4 m/s , - 1 m/s
Explanation:
mass of first block, m1 = 1 kg
mass of second block, m2 = 4 kg
initial velocity of first block, u1 = 5 m/s
initial velocity of second block, u2 = 0 m/s
Let v is the velocity of centre of mass.
The formula used for the velocity of centre of mass is given by
[tex]v=\frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}[/tex]
[tex]v=\frac{1\times 5+4\times 0 }{5}[/tex]
v = 1 m/s
So,
the velocity of first block with respect to centre of mass frame = 5 - 1
= 4 m/s
the velocity of second block with respect to centre of mass frame = 0 - 1
= -1 m/s
Answer:
The initial velocities of the 1-kg and the 4-kg blocks in the center-of-mass reference frame is 4 m/s and (-1 m/s) respectively.
Explanation:
Given that,
Mass of first block, m = 1 kg
Mass of another block, m' = 4 kg
Initial speed of 1 kg block, u = +5 m/s
Initial speed of 4 kg block, u' = 0 (it was at rest)
Let V is the velocity of the center of mass. It is given by :
[tex]V=\dfrac{mu+m'u'}{m+m'}\\\\V=\dfrac{1\times 5+4\times 0}{1+4}\\\\V=1\ m/s[/tex]
So, the speed of center of mass is 1 m/s
Now, the initial velocity of 1 kg block in the center-of-mass reference frame is, 5 m/s - 1 m/s = 4 m/s
the initial velocity of 4 kg block in the center-of-mass reference frame is, 0 m/s - 1 m/s = -1 m/s
Hence, the initial velocities of the 1-kg and the 4-kg blocks in the center-of-mass reference frame is 4 m/s and (-1 m/s) respectively.
