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Mrrafil [7]
3 years ago
12

A car traveling 13m/s to the north collideswith a car traveling 19m/s to the south. Each car has a mass of 1250 kg. If the syste

m is defined as the two cars, what is the change in momentum?
Physics
2 answers:
GalinKa [24]3 years ago
5 0
Let P1=13m/s(1250kg)=16250 kg m/s  (positive velocity is north here)
      P2=-19m/s(1250kg)=-23750 kg m/s
Total system momentum is P1 + P2 = 16250 - 23750 = -7500 (negative momentum is south pointing).
Due to the conservation of momentum, this is also the final momentum and we see the two cars have a mass of 1250 + 1250 = 2500kg and have a velocity
-7500 = 2500kg * Vfinal
Final = -3m/s (moving southward)
Each car's momentum changes from the initial value it had to this final value, but the total system momentum does not change, due to the conservation law.
mestny [16]3 years ago
4 0

Answer:

0

Explanation:

Other guy worded it weird but he’s right

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2 years ago
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia
Oliga [24]

Answer with Explanation:

We are given that

m_1=0.3 kg

m_2=0.4 kg

Spring constant,k=200 N/m

h=0.250 m

a.Speed of ham before collision

u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s

Collision is inelastic

According to law of conservation of momentum

m_1u_1=(m_1+m_2)v

0.3\times 2.2=(0.3+0.4)v

v=\frac{0.3\times 2.2}{0.7}=0.94 m/s

Kinetic energy of masses=Spring potential energy at maximum distance

\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2

A^2=\frac{(m_1+m_2)v^2}{k}

A=\sqrt{\frac{(m_1+m_2)v^2}{k}}

Substitute the values

A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm

1 m=100 cm

b.Time period,T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}

T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s

7 0
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