Answer:
3.88m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and 2 are the initial velocities
v is the final velocity
Given
m1 = 64kg
u1 = 4.2m/s
m2 = 25kg
u2 = 3.2m/s
Required
Final velocity v
Substitute the given values into the formula
64(4.2)+25(3.2) = (65+25)v
268.8+80 = 90v
348.8 = 90v
v = 348.8/90
v = 3.88m/s
Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s
The correct answer to the question is : C) The horizontal momentum and the vertical momentum are both conserved.
EXPLANATION :
Before coming into any conclusion, first we have to understand the law of conservation of momentum.
As per the law of conservation of momentum, the total linear as well as angular momentum of an isolated system is always conserved . The law of conservation of energy is a universal fact.
Hence, during any type of collision, the total momentum is always conserved.
Hence, the total horizontal momentum as well as total vertical momentum are always conserved during both elastic as well as inelastic collision.
Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
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