A horse pulls a cart along a road with a force of 550 lbs. If the horse does 2,674,100 ftlbs of work by the time it stops, how f
2 answers:
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Answer:
Magnitude of the velocity = 16.82 m/s
Angular direction, θ = 52.41°
Explanation:
As given ,
mass of bullet, m₁= 50g = 0.05 kg
speed of bullet , u₁ = 600 m/s
mass of the block , m₂ = 4 kg
speed of the block before collision , u₂ = 12 m/s
direction , θ = 30°
Now,
Assume that the combined velocity of bullet and block after collision = v
and the direction = θ
Now, from the conservation of momentum in x - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ
where v = final velocity after collision
u₁ = initial velocity of bullet before collision = 0
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 cos(30° )
m₂ = mass of block before collision
m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ
⇒ 0 + 4(6√3) = 4.05 vₓ
⇒24√3 = 4.05 vₓ
⇒vₓ = 10.26 m/s
Now, from the conservation of momentum in y - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ )
where v = final velocity after collision
u₁= initial velocity of bullet before collision = 600
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 sin(30° )
m₂= mass of block before collision
m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 )
⇒ 30 + 4(6) = 4.05
⇒30 +24 = 4.05
⇒54 = 4.05
⇒ = 13.33 m/s
Now, the magnitude of the velocity = √vₓ² + ² = √(10.26)² + (13.33)²
= √105.26 + 177.68
= √282.95 = 16.82
The angular direction, θ = (
) =
(
) =
(1.299) = 52.41°