The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
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- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum





Answer:
v1=18.46m/s
v2=29.8cm/s
Explanation:
We know that

the equation of the motion is

we can calculate w by using

Hence, we have that

the speed will be

hope this helps!
Answer is x hopefully this helps your stupid head
288.51 N is the magnitude of the force that the beam exerts on the hi.nge.
Given
Mass 0f beam = 40 Kg
The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N
Angle between the beam and cable is = 90°
Angle between beam and the horizontal component = 31°
As the system of the beam, hi_nge and cable are in equilibrium.
The magnitude of the force that the beam exerts on the hi_nge can be calculated by -
F =The horizontal component of force + the vertical component of force
F = 86.62 N + 40 × 9.8 × sin 31°
F =86.62 N + 201.89 N
F = 288.51 N
Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.
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