Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
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