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Elena L [17]
4 years ago
13

In outer space a five KG object moving in to the right at 4M/S and a two KG object moving to the left at 3MS collide and stick t

ogether their speed after the collision is
Physics
1 answer:
Anettt [7]4 years ago
7 0

Answer:

Combined speed, V = 2 m/s

Explanation:

Given that,

Mass of object 1, m = 5 kg

Speed of object 1, u = 4 m/s (due right)

Mass of object 2, m' = 2 kg

Speed of object 2, u' = -3 m/s (due left)

<em>Let V is the speed of the objects after the collision. Here, both objects stick to each other. Therefore, the momentum remains conserved. So,</em>

mu+m'u'=(m+m')V\\\\V=\dfrac{mu+m'u'}{(m+m')}\\\\V=\dfrac{5\times 4+2\times (-3)}{(5+2)}\\\\V=2\ m/s

So, the combined speed of the masses will be 2 m/s.

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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
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Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

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The initial potential energy of the spring is given by the equation:

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K_{f}=\frac{1}{2}mv_{f}^{2}

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=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

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