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what was the final momentum?
what was the initial velocity of the 0.2 kg cart?
if the collision lasted 0.02 s, what force did each cart exert on the other?​

What Was The Final Momentumwhat Was The Initial Velocity Of The 02 Kg Cartif The Collision Lasted 002 S What Force Did Each Cart Exert On The Other class=

Sagot :

Leenaidn

Hello!

Recall the equation for momentum:


[tex]\huge\boxed{ p = mv}[/tex]

p  = linear momentum (kgm/s)

m = mass (kg)

v = velocity (m/s)

Part 1:

We can solve for the total momentum using the above equation. Let m1 represent the 0.2 kg cart, and m2 represent the 0.4 kg cart.

[tex]p = m_1v_1' + m_2v_2'[/tex]

Since they move off together:

[tex]p = v_f(m_1 + m_2) = 0.2(0.2 + 0.4) = \boxed{0.12 kg\frac{m}{s}}[/tex]

Part 2:

Using the conservation of momentum:

[tex]m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\\\\[/tex]

m2 was initially at rest, so:


[tex]m_1v_1 + m_2(0) = 0.12\\\\0.2(v_1) = 0.12\\\\v_1 = \frac{0.12}{0.2} = \boxed{0.6 \frac{m}{s}}[/tex]

Part 3:

We can calculate the force by first calculating the impulse exerted on the carts.

Recall the equation for impulse:

[tex]\large\boxed{I = m\Delta v = m(v_f - v_i}}[/tex]

We can use either cart, but for ease, we can use the 0.4 cart that starts from rest.

Thus:
[tex]I = 0.4(0.2 - 0) = 0.08 kg\frac{m}{s}[/tex]

Now, calculate force with the following:
[tex]\large\boxed{I = Ft, F = \frac{I}{t}}[/tex]

Plug in the values:
[tex]F = \frac{0.08}{0.02} = \boxed{4 N}[/tex]

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