Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m - 25 i m + 2 m v
2 v=25 j + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s
Answer:
Part a)
Part b)
North of East
Explanation:
Speed of train towards East = 60 km/h
displacement towards East is given as
now it turns towards 50 degree East of North
so its distance is given as
then finally it moves towards west for 50 min
Now the total displacement of the train is given as
now total time duration of the motion is given as
now average velocity is given as
Part a)
magnitude of the average velocity is given as
Part b)
Direction of the velocity is given as
North of East
Answer:
(9.64 +- 0.86) m/s^2
Explanation:
The generic motion equation for constant acceleration is
Where
X0: initial position
v0: initial speed
a: acceleration
t: time
If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:
And the acceleration can be obtained as:
Where x is the distance fallen and a = g.
So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate
For the uncertainty we have to calculate the relative uncertainties first
For the distance (100 * 0.3)/100 = 0.3%
For the time (100 * 3)/144 = 2.08%
For multiplications or divisions the relative uncertainties are added
0.3% + 2.08% + 2.08% = 4.46%
We convert this into absolute uncertainty:
(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)
Finally, this is multiplied by a constant scalar, so:
2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)
We convert the units
0.86 m/(s^2)
And the measurement is (9.64 +- 0.86) m/s^2
A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.