Answer:
1.5 m/s
Explanation:
Momentum is conserved and conservation of momentum is
p₁ + p₂ = p'₁ + p'₂
or
m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂
In our problem, after collision v'₁ will be equal to v'₂.
Since objects are identical m₁ = m₂
m(v₁+ v₂) = 2m x v'₁
(2m/s + 1m/s) = 2v'₁
v'₁ = v'₂ = 1.5 m/s
Answer:
a) Maximum speed = 25.28 m/s
b) Total time = 27.27 s
c) Total distance traveled = 402.43 m
Explanation:
a) Maximum speed is obtained after the end of acceleration
v = u + at
v = 13.5 + 1.9 x 6.2 = 25.28 m/s
Maximum speed = 25.28 m/s
b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.
v = u + at
0 = 25.28 - 1.2 t
t = 21.07 s
Total time = 6.2 + 21.07 = 27.27 s
c) Distance traveled for the first 6.2 s
s = ut + 0.5 at²
s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m
Distance traveled for the second 21.07 s
s = ut + 0.5 at²
s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m
Total distance traveled = 120.22 + 282.21 = 402.43 m
3 is false 2 is true and the rest true
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
