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4 years ago

Physics !!!!!!!!!!!!!!!!!

Physics

1 answer:

madreJ [45]4 years ago

3 0

Answer:

1) v = -0.89 m/s

2) v = 36.5 m/s

Explanation:

1) Given

The mass of the boy, M = 45 Kg

The mass of the skateboard, m = 2 Kg

The initial velocity of the boy and skateboard is zero

According to the law of conservation of linear momentum

MV + mv = 0

V = - mv/M

Substituting the given values in the above equation

V = -2 x 20/ 45

= -0.89 m/s

Hence, the velocity of the boy, v = -0.89 m/s

2) Given

The mass of the boy, M = 47 Kg

The mass of the skateboard, m = 8 Kg

The initial combined velocity of the boy and skateboard, u = 4.2 m/s

The boy jumped off from the skateboard with velocity, V = -1.3 m/s

According to the law of conservation of momentum,

MV + mv = (M + m)u

v = [(M + m)u - MV] / m

Substituting the given values,

v = [(47 + 8)4.2 - 47(-1.3)] / 8

= 36.5 m/s

Hence, the velocity of the skateboard, v = 36.5 m/s

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Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

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For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

Second stone

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

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