Question 10 of 10

The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, wha

svp [43]

Given Information:

Initial speed of rock = vi = 30 m/s

escape speed of the asteroid = ve = 24 m/s

Required Information:

final speed of rock = vf = ?

Answer:

vf = 18 m/s

Explanation:

As we know from the conservation of energy

KEf + Uf = KEi + Ui

Where KE is the kinetic energy and U is the potential energy

0 + 0 = ½mve² - GMm/R

When escape speed is used, KEf is zero due to vf being zero. Uf is zero because the object is very far away from mass M, therefore, the equation becomes

GMm/R = ½mve²

m cancels out

GM/R = ½ve²

GM/R = ½(24)²

GM/R = 288

KEf + Uf = KEi + Ui

½mvi² + 0 = ½vf² - GMm/R

m cancels out

½vi² = ½vf² - GM/R

Substitute the values

½(30)² = ½vf² - (288)

½vf² = 450 - 288

vf² = 2(162)

vf = √324

vf = 18 m/s

Therefore, the final speed of the rock is 18 m/s

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4 years ago

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A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$