Question

Consider a glancing collision between two identical spheres of 0.016 kg, with one of the spheres initially at rest. Initially the incoming projectile has a velocity of 1.50 m/s to the right and after the collision we observe that this sphere leaves the collision region with a velocity of 0.80 m/s at an angle of 25 to its initial direction. (Assume that the +x-axis is to the right and the +y-axis is up along the page. Angles are measured counterclockwise from the +x-axis. Do not assume this collision is elastic.)

A)Find the x-component of the velocity of the second sphere.

B)Find the y-component of the velocity of the second sphere.

Answer 1

(a) The x-component of the second sphere's velocity is 0.775 m/s

(b) The y-component of the second sphere's velocity is -0.338 m/s

Step-by-step explanation:

Conservation of Linear Momentum

The principle of the conservation of linear momentum states that the total momentum of a closed system of particles is conserved while no external forces act on any part of it, regardless of the interaction between the particles. The momentum is a vector, thus we must analyze both axes in the calculations.

The momentum can be computed as

`\vec p=m.\vec v`

where m is the mass of the particle, and
`\vec v` is its velocity. In a system of two particles, the total initial momentum is

`\vec p_o=m_1.\vec v_1+m_2.\vec v_2`

And the final momentum is

`\vec p_1=m_1.\vec v_1'+m_2.\vec v_2'`

Since the total momentum is conserved

`m_1.\vec v_1'+m_2.\vec v_2'=m_1.\vec v_1+m_2.\vec v_2`

According to the conditions of the problem, both masses are identical and sphere 2 is initially at rest (v2=0), thus

`m.\vec v_1'+m.\vec v_2'=m.\vec v_1`

Simplifying by m

`\vec v_1'+\vec v_2'=\vec v_1\text{ ......[1]}`

We know that the first sphere has a velocity of 1.5 m/s to the right. This means that the vertical component of v1 is 0:

`\vec v_1=<1.5,0>`

We also know after the collision this same sphere travels at 0.8 m/s at an angle of 25°. The components of this velocity are

`\vec v_1'=<0.8cos25^o,0.8sin25^o>`

(a) Solving the equation [1] for v2

`\vec v_2'=\vec v_1-\vec v_1'`

`\vec v_2'=<1.5,0>-<0.8cos25^o,0.8sin25^o>`

`\vec v_2'=<1.5,0>-<0.725,0.338>`

`\vec v_2'=<0.775,-0.338>`

The x-component of the second sphere's velocity is 0.775 m/s

(b) The y-component of the second sphere's velocity is -0.338 m/s