Question

A cart with mass m1 = 3.2 kg and initial velocity of v1,i = 2.1 m/s collides with another cart of mass M2 = 4.3 kg which is initially at rest in the lab frame. The collision is completely elastic, and the wheels on the carts can be treated as massless and frictionless. What is the velocity of m1 in the center of mass frame after the collision?

Answer 1

Answer:

-1.204 m/s

Explanation:

$m_1$ = Mass of first cart = 3.2 kg

$m_2$ = Mass of second cart = 4.3 kg

$u_1$ = Initial Velocity of first cart = 2.1 m/s

$u_2$ = Initial Velocity of second cart = 0 m/s

$v_c$ = Velocity of center of mass

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v_c\\\Rightarrow v_c=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow  v_c=\frac{3.2\times 2.1 + 4.3\times 0}{3.2 + 4.3}\\\Rightarrow v_c=0.896\ m/s$

Velocity of $m_1$

$V_m_1=2.1-0.896=1.204\ m/s$

When considering the center of mass frame $V_m_1=-1.204\ m/s$

Answer 2

The velocity of $m_1$ in the center of mass frame after the collision is equal to 1.204 m/s.

Given the following data:

• Mass of cart 1 = 3.2 kg.

• Mass of cart 2 = 4.3 kg.

• Initial velocity of cart 1 = 2.1 m/s.

• Initial velocity of cart 2 = 0 m/s.

What is an elastic collision?

An elastic collision can be defined as a type of collision between two objects wherein the total kinetic energy of the two (2) objects is conserved (remains the same) after the collision.

How to calculate the velocity of $m_1$ in the center of mass frame.

Mathematically, an elastic collision for the two carts is given by this formula:

$m_1u_1+m_2u_2 =(m_1+m_2)v_c\\\\V_c=\frac{m_1u_1+m_2u_2}{m_1+m_2} \\\\V_c=\frac{(3.2 \times 2.1)+4.3 \times 0}{3.2+4.3}\\\\V_c=\frac{6.72}{7.5} \\\\V_c=0.896\;m/s$

For the velocity of mass 1, we have:

$V_{m1}=U_1 -V_c \\\\V_{m1}=2.1 -0.896\\\\V_{m1}=1.204\;m/s.$

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