Question

A cylindrical metal specimen 15.0 mm (0.59 in.) in diameter and 150 mm (5.9 in.) long is to be subjected to a tensile stress of 50 Mpa (7250 psi); at this stress level, the resulting deformation will be totally elastic.

a) If the elongation must be less than 0.072 mm (2.83 x 10−3−3 in.), which of the metals is Table 6.1 are suitable candidates? Why?
b) If, in addition, the maximum permissible diameter decrease is 2.3 X 10−3−3 mm (9.1 x 10−5−5 in.) when the tensile stress of 50 Mpa is applied, which of the metals that satisfy the criterion in part (a) are suitable candidates? Why?

Answer 1

We can solve this problem by applying the principle of conservation of momentum and using vector addition to determine the final velocities of the two balls.

According to the principle of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. In other words, the sum of the initial momenta of the two balls must be equal to the sum of their final momenta.

Let's assume that the positive x-axis is to the right and the positive y-axis is upwards. Then, the initial momentum of the system can be expressed as:

p_initial = m_green * v_green,i + m_red * v_red,i

where m_green and m_red are the masses of the green and red balls, respectively, and v_green,i and v_red,i are their initial velocities.

Substituting the given values, we get:

p_initial = (10 kg) * (25 m/s) + (15 kg) * (0 m/s) = 250 kg m/s

After the collision, the green ball moves at a 35 degree angle to the left of its original direction, which means its velocity has both horizontal and vertical components. We can resolve the green ball's final velocity into x- and y-components as follows:

v_green,x = v_green,f * cos(35°)

v_green,y = v_green,f * sin(35°)

where v_green,f is the magnitude of the green ball's final velocity.

Similarly, we can resolve the red ball's final velocity into x- and y-components as follows:

v_red,x = v_red,f * cos(55°)

v_red,y = v_red,f * sin(55°)

where v_red,f is the magnitude of the red ball's final velocity.

Since momentum is conserved in both the x- and y-directions, we can write two equations:

m_green * v_green,i = m_green * v_green,x + m_green * v_green,y + m_red * v_red,x + m_red * v_red,y (in the x-direction)

0 = m_green * v_green,y - m_red * v_red,y (in the y-direction)

Substituting the resolved components of the final velocities, we get:

(10 kg) * (25 m/s) = (10 kg) * v_green,f * cos(35°) + (10 kg) * v_green,f * sin(35°) + (15 kg) * v_red,f * cos(55°) + (15 kg) * v_red,f * sin(55°)

0 = (10 kg) * v_green,f * sin(35°) - (15 kg) * v_red,f * sin(55°)

Simplifying and solving for v_red,f, we get:

v_red,f = [(10 kg) * (25 m/s) - (10 kg) * v_green,f * cos(35°) - (10 kg) * v_green,f * sin(35°)] / [(15 kg) * cos(55°) + (15 kg) * sin(55°)]

Similarly, we can solve for v_green,f:

v_green,f = [(10 kg) * v_green,f * cos(35°) + (10 kg) * v_green,f * sin(35°) - (15 kg) * v_red,f * sin(55°)] / (10 kg)

Simplifying further, we get:

v_red,f = (250 - 10 v_green,f) / (15 cos(55°) + 15 sin(55°)) ≈ 8.41 m/s

v_green,f = (2/3) * (25 m/s) / (cos(35°)