Question

A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?

Answer 1

Answer:

0.0158m/s

Explanation:

Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;

m1u1 + m2u2 =  m1v1 + m2v1

m1 and m2 are the masses of the object

u1 and u2 are the initial velocities

v1 and v2 are the final velocities

Given

m1 = 300g  = 0.3kg

u1 = 1.10m/s

m2 = 4.00kg

u2 = 0m/s (at rest)

v1 = 0.890

v2 = ?

Substitute the given values into the formula;

0.3(1.10) + 0 = 0.3(0.89) + 4v2

0.33 = 0.267 + 4v2

0.33-0.267 = 4v2

0.063 = 4v2

v2 = 0.063/4

v2 = 0.0158m/s

Hence the speed of the large cart after the collision is 0.0158m/s

Answer 2

The speed of the large cart will be "0.0158 m/s".

According to the question:

Mass,

• $m_1 = 300 \ g = 0.3 \ kg$
• $m_2 = 4.00 \ kg$

Final velocity,

• $u_1 = 1.10 \ m/s$
• $u_2 = 0 \ m/s$

Initial velocity,

• $v_1 = 0.890 \ m/s$
• $v_2 = \ ?$

By using the law of conservation, we get

→ $m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1$

By substituting the values, we get

→ $0.3(1.10)+0 = 0.3(0.89)+4v_2$

→              $0.33=0.267+4 v_2$

   $0.33-0.267 = 4 v_2$

              $0.063 =4v_2$

                   $v_2 = \frac{0.063}{4}$

                       $= 0.0158 \ m/s$

Thus the response above is appropriate.

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