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2 years ago

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A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon

. What is the final speed of the child and wagon?

1 answer:

Shalnov [3]2 years ago

4 0

Answer:

The final velocity is $u_f = 3.44 \ m/s$

Explanation:

From the question we are told that

The mass of the child is $m_1 = 20 \ kg$

The initial speed of the child is $u_1 = 3.3 \ m/s$

The mass of the wagon is $m_w = 4.0 \ kg$

The initial speed of the wagon is $u_w = 3.3 \ m/s$

The mass of the ball is $m_2 = 1.0 \ kg$

The initial speed off the ball is $u_2 = 3.3 \ m/s$

Generally the initial speed of the system (i.e the child , wagon , ball) is

$u_1 = u_w = u_2 = u =3.3 \ m/s$

Generally from the law of linear momentum conservation

$p_i = p_f$

Here $p_i$ is the momentum of the system before the ball is dropped which is mathematically represented as

$p_i = ( m_1 + m_2 + m_3 ) * u$

=> $p_i = ( 20 + 4 + 1 ) * 3.3$

=> $p_i = 82.5 \ kg \cdot m/s$

and

$p_f$ is the momentum of the system after the ball is dropped which is mathematically represented as

$p_f = ( m_1 + m_w ) * u_f$

=> $p_i = ( 20 + 4 ) * u_f$

So

$82.5 = 24 * u_f$

=> $u_f = 3.44 \ m/s$

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As per above given data

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now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

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now the acceleration is given as

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Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

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R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

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