Answer:
v = 2.57 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. When it is at an angle of 30º
Em₀ = K + U = ½ m v₁² + m g y₁
final point. Lowest position
Em_f = K = ½ m v²
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v₁² + m g y₁ = ½ m v²
Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle
cos 30 = L ’/ L
L ’= L cos 30
y₁ = L -L '
y₁ = L- L cos 30
we substitute
½ m v₁² + m g L (1- cos 30) = ½ m v²
v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]
let's calculate
v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]
v = 2.57 m / s
A 64.0-kg person holding two 0.900-kg bricks stands on a 2.00-kg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 19.0 m/s. What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?
Answer:
0.518 m/s
Explanation:
To solve this question, we would use the Law of conservation of momentum.
It is stated that the skateboard and the person were at rest initially, this means that their initial momentum is 0. Since they aren't moving.
And from the law if conservation of momentum, we know that initial momentum must be equal to final momentum.
The final momentum of the bricks will be 2* 0.9 * 19 = 34.2 kg m/s.
This means that the momentum of the person and that of the skateboard has to be 34.2 kg m/s also, although, it will be in the opposite direction.
Mass of the person plus that of the skateboard is
64 + 2 = 66 kg,
If we divide the momentum of the person and that of the skateboard by the mass of the both of them, we get the speed at which they are moving after the bricks are thrown
Recoil speed = (34.2 kg m/s)/(66 kg) = 0.518 m/s.
The recoil speed of the person is 0.518 m/s
Calculation of the recoil speed of the person:
As per the conservation of momentum, the initial momentum should be equivalent to the final momentum
So, here the final momentum should be
= 2* 0.9 * 19
= 34.2 kg m/s.
Now the mass of the person should be
= 64 + 2
= 66 kg,
Now the recoil speed should be
Recoil speed = (34.2 kg m/s)/(66 kg)
= 0.518 m/s.
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