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liberstina [14]
3 years ago
9

A small toy cart equipped with a spring bumper rolls toward a wall with a speed of v . The cart rebounds from the wall, with the

same speed v . The sketch below shows the initial velocity vector v⃗i , final velocity vector v⃗f , and the initial momentum vector p⃗i . Using the initial momentum vector as a basis, draw the change in momentum vector Δp⃗ for the cart.
The orientation and length of your vector will be graded. The location will not be graded.

Physics
1 answer:
anastassius [24]3 years ago
5 0

Using the initial momentum vector as a basis, the change in momentum vector Δp for the cart is drawn as shown in the attachment.

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial speed of cart = v_i = v

Final speed of cart = v_f = v

<u>Unknown:</u>

The change in momentum of cart = I  = ?

<u>Solution:</u>

I = \Delta p

I = p_f - p_i

I = mv_f - mv_i

I = m ( v_f - v_i )

I = m ( -v - v )

I = m ( -2v )

I = -2mv

I = -2p_i

\texttt{ }

<em>From the results above, we can conclude that the change in momentum vector Δp is twice the initial momentum vector p_i but in opposite direction.</em>

The vector <em>Δp could be drawn as shown </em><em>in the attachment.</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

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Explanation:

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initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

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let final velocity of the bullet-wooden block system after collision = v₂

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Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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Answer:

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