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Furkat [3]
3 years ago
6

Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b

all, he and the ball move on the skateboard. How fast do Jocko and the ball move after he catches the ball?
Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

Mass if Jocko, m = 60 kg

Mass of the ball, m' = 20 kg

Speed of the ball, v = 3 m/s

Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

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Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

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m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

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Explanation
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