Question

A ball with a mass of 0.585 kg is initially at rest. It is struck by a second ball having a mass of 0.420 kg , initially moving with a velocity of 0.270 m/s toward the right along the x axis. After the collision, the 0.420 kg ball has a velocity of 0.220 m/s at an angle of 36.9 ∘ above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.
Required:
a. What is the magnitude of the velocity of the 0.605kg ball after the collision?
b. What is the direction of the velocity of the 0.605kg ball after the collision?
c. What is the change in the total kinetic energy of the two balls as a result of the collision?

Answer 1

Answer:

a) $(v_1)=0.3989m/s$

b) $\theta_1=80.5 \textdegree$

c) $K.E=0.036J$

Explanation:

From the question we are told that:

Initial speed of 1st ball $u_{1}=0 m/s$

Mass of 1st ball $m_1=0.585kg$

Mass of 2nd ball $m_2=0.420kg$

Initial speed of 2nd ball $u_{2}=0.270 m/s$

Final speed of 2nd ball $v_{2}=0.220 m/s$

Angle of collision $\angle=36.9 \textdegree$

a)

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2$

The final velocity $v_1$ is given as

$0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2$

$(v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}$

$(v_1)^2=0.1591$

$(v_1)=0.3989m/s$

b)

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2$

$0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree$

$cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}$

$cos\theta_1=0.1656$

$\theta_1=80.5 \textdegree$

c)

Generally the equation for kinetic energy is mathematically given by

$K.E=\frac{1}{2} mv^2$

1st Ball

$K.E=\frac{1}{2} (0.585)(0.3989)^2$

$K.E=0.0465J$

2nd ball

$K.E=\frac{1}{2} (0.420)(0.220)^2$

$K.E=0.101J$

Therefore the  change in the total kinetic energy of the two balls as a result of the collision is

$0.101-0.0465$

$K.E=0.036J$