Answer:
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Explanation:
Mass of object 1 , m₁ = 300 g = 0.3 kg
Mass of object 2 , m₂ = 400 g = 0.4 kg
Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s
Initial velocity of object 2 , v₂ = 3.00j m/s
Mass of composite = 0.7 kg
We need to find final velocity of composite.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s
Final momentum = 0.7 x v = 0.7v kgm/s
Comparing
1.5 i + 0.24 j = 0.7v
v = 2.14 i + 0.34 j
Magnitude of velocity
Direction,
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
Because the atoms and molecules all have different properties
Answer:
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Explanation:
