Question

phy 1010/1080 - activity 6 - 7
chapter 6 - momentum & impulse
chapter 7 - rotational motion
chapter 6 problems:
for all problems, draw a diagram if applicable and show your work unless otherwise noted.

1. two rubber balls of masses ( m ) and ( 2m ) are heading directly towards one another on a frictionless surface. if the ball of mass ( m ) is traveling at a velocity of ( v ), how fast must the ball of mass ( 2m ) be traveling for both to ricochet at equal speeds after collision? assume that this is a perfectly elastic collision.
2. two clay balls of masses ( m ) and ( 2m ) are heading directly towards one another on a frictionless surface. if the ball of mass ( m ) is traveling at a velocity of ( v ), how fast must the ball of mass ( 2m ) be traveling for there to be no motion after the collision? assume that this is a perfectly inelastic collision.
3. a baseball of mass ( 0.145 mathrm{~kg} ) is thrown at ( 45.0 mathrm{~m} / mathrm{s} ) and is batted away at ( 50.0 mathrm{~m} / mathrm{s} ) by the batter with a ( 1.00 mathrm{~kg} ) bat after being in contact with the bat for ( 1.00 mathrm{~ms} ). what is the acceleration of the batters bat?

Explanation:

Step1: Use conservation of momentum

For a perfectly elastic collision, the law of conservation of momentum states \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Let the velocity of the ball with mass \(m\) be \(v\) (initial velocity, \(u_1 = v\)), mass \(m_1=m\), mass \(m_2 = 2m\), let the initial velocity of the ball with mass \(2m\) be \(u_2\). After collision, let the speed of both be \(s\). Since they ricochet, \(v_1=-s\) and \(v_2 = s\).

$$mv+2mu_2=-ms + 2ms$$

Step2: Use conservation of kinetic energy

For a perfectly elastic collision, the law of conservation of kinetic energy states \(\frac{1}{2}m_1u_1^{2}+\frac{1}{2}m_2u_2^{2}=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}\).

$$\frac{1}{2}mv^{2}+\frac{1}{2}(2m)u_2^{2}=\frac{1}{2}m(-s)^{2}+\frac{1}{2}(2m)s^{2}$$

$$mv^{2}+2mu_2^{2}=ms^{2}+2ms^{2}$$

$$v^{2}+2u_2^{2}=3s^{2}$$

From the momentum equation \(mv+2mu_2=ms\), we can express \(s=v + 2u_2\). Substitute \(s\) into the kinetic - energy equation:

$$v^{2}+2u_2^{2}=3(v + 2u_2)^{2}$$

$$v^{2}+2u_2^{2}=3(v^{2}+4u_2v+4u_2^{2})$$

$$v^{2}+2u_2^{2}=3v^{2}+12u_2v + 12u_2^{2}$$

$$0=2v^{2}+12u_2v+10u_2^{2}$$

$$0 = v^{2}+6u_2v + 5u_2^{2}$$

Factor the quadratic equation: \((v + u_2)(v+5u_2)=0\). We get \(u_2=-v\) or \(u_2 =-\frac{v}{5}\). We take \(u_2=\frac{v}{5}\) (magnitude, considering direction in the momentum equation).

Answer:

The ball of mass \(2m\) must be traveling at a speed of \(\frac{v}{5}\)