A bullet is having a mass of 10g and moving with a speed of 1.5 m/s penetrates a thick wooden plank of mass 99g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.
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3
Given,
m₁ = 10g = 0.01kg
m₂ = 99g = 0.099kg
u₁ = 1.5m/s
u₂ = 0
v₁ = v₂ = v
Now,
m₁u₁ + m₂u₂ = m₁v + m₂v
(0.01)(1.5) + (0.099)(0) = 0.01v + 0.099v
0.015 = 0.109v
v = 0.015/0.109
v = 15/109 m/s
Therefore they moved together with a speed of 15/109 m/s
m₁ = 10g = 0.01kg
m₂ = 99g = 0.099kg
u₁ = 1.5m/s
u₂ = 0
v₁ = v₂ = v
Now,
m₁u₁ + m₂u₂ = m₁v + m₂v
(0.01)(1.5) + (0.099)(0) = 0.01v + 0.099v
0.015 = 0.109v
v = 0.015/0.109
v = 15/109 m/s
Therefore they moved together with a speed of 15/109 m/s
Answered by
2
Initial momentum of system = mu
= 10 × 10^-3 kg × 1.5 m/s
= 0.015 kgm/s
Final momentum of system = (m + M)v
= [(10 + 99) × 10^-3 kg] × v
= 0.109v
According to law of conservation of momentum
Initial momentum of system = Final momentum of system
0.015 = 0.109v
v = 0.1376 m/s
v = 13.76 cm/s
Their combine velocity is 13.76 cm/s
= 10 × 10^-3 kg × 1.5 m/s
= 0.015 kgm/s
Final momentum of system = (m + M)v
= [(10 + 99) × 10^-3 kg] × v
= 0.109v
According to law of conservation of momentum
Initial momentum of system = Final momentum of system
0.015 = 0.109v
v = 0.1376 m/s
v = 13.76 cm/s
Their combine velocity is 13.76 cm/s
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