Answer:
Explanation:
Mass of first cart M1=2.4kg
Velocity of first cart U1=4.1m/s
Mass of second cart M2=1.7kg
Second cart is initially at rest U2=0
After an instant, the velocity of the second cart is U2=-2.8m/s
Now after collision the two cart move together with the same velocity I.e inelastic collision
Using conservation of momentum
Momentum before collision, = momentum after collision
M1U1 + M2U2 = (M1+M2)V
2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V
9.84 - 4.76 = 4.1V
5.08=4.1V
V=5.08/4.1
V=1.24m/s
The momentum of the two cart at that instant is
M1U1+M2U2
2.4×4.1 + 1.7× -2.8
9.84 - 4.76
5.08kgm/s
So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s
k = 5.29
a = 0.78m/s²
KE = 0.0765J
<u>Explanation:</u>
Given-
Mass of air tracker, m = 1.15kg
Force, F = 0.9N
distance, x = 0.17m
(a) Effective spring constant, k = ?
Force = kx
0.9 = k X0.17
k = 5.29
(b) Maximum acceleration, m = ?
We know,
Force = ma
0.9N = 1.15 X a
a = 0.78 m/s²
c) kinetic energy, KE of the glider at x = 0.00 m.
The work done as the glider was moved = Average force * distance
This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0
As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)
Work = Kinetic energy
KE = 0.450 * 0.17
KE = 0.0765J
Answer:
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