Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
if it moves 7m/s, that means every second it goes 7m. Now we just multiply by the time (20 seconds) and end up with 140m.
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Answer:
-2.478
0.379
11.14
24.78
Explanation:
Angular frequency of spring in harmonic motion is given by?
ω = √(k/m)
ω = √(10/2.2)
ω = √4.54
ω = 2.13 s^-1
If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine
x(t) = -A * cos(ωt)
x(t) = -2.48 * cos(2.13 * 1)
x(t) = -2.48 * 0.9993
x(t) = -2.478
Velocity and acceleration are 1st and 2nd derivative of position
b)
v(t) = Aω * sin(ωt)
v(t) = 2.48 * 2.13 * sin(2.13 * 1)
v(t) = 5.282 * sin2.13
v(t) = 5.282 * 0.03717
v(t) = 0.379 m/s
c)
a(t) = Aω^2 * cos(ωt)
a(t) = 2.48 * 2.12² * cos(2.13 * 1)
a(t) = 2.48 * 4.494 * cos2.13
a(t) = 11.15 * 0.9993
a(t) = 11.14 m/s²
d)
F = -k * x(t)
F = -10 * -2.478
F = 24.78 N
Due to the law of conservation of momentum, the force exerted on the mallet is equal and opposite to the force exerted on the ball, so the answer is C.