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nlexa [21]
2 years ago
9

A bullet of mass 0.01 kg is fired in to a sand bas of mass 0.49 kg from a tree. The Sand bag with the bullet embedded in to it s

wings a way at lo m/s find a The momentum after the Collision b, The momentum befor the Collision the velocity of the bullet ​
Physics
2 answers:
makvit [3.9K]2 years ago
8 0

Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.

<h3>Linear momentum</h3>

The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.

Momentum before collision for the bullet = 0.01 kg × v

Momentum before collision for the bag = 0

Momentum after collision for the bag and bullet = (0.01 kg  +  0.49 kg) 10 = 5 Kgms-1

The velocity of the bullet before collision =  0.01 kg × v + 0 =  5 Kgms-1

v = 5 Kgms-1/0.01 kg

v = 500 ms-1

Learn more about momentum: brainly.com/question/904448

antoniya [11.8K]2 years ago
7 0

Answer:

(0.01kg \times 0.49)10ms = 5kgms

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

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u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
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