I've been stuck on this questions for 2 days
A 7.2-kg bowling ball traveling at a speed of 3.5 m/s collides with a 1.2-kg bowling pin at rest. After the collision, the pin moves with a speed of 3.0 m/s at an angle of 60° with respect to the original direction of motion of the bowling ball. What is the final velocity of the ball (magnitude and direction)? Is the collision approximately elastic?
momentum is conserved in both the x and y directions
the initial momentum is all in the x direction -- the moving ball
PMy = 3.0 m/s * sin(60º) * 1.2 kg
BVy = PMy / 7.2
PMx = 3.0 m/s * cos(60º) * 1.2 kg
BVx = 3.5 - (PMx / 7.2)
use Pythagoras to find BV
To solve this problem, we can use the principle of conservation of momentum and the law of conservation of kinetic energy.
Step 1: Calculate the initial momentum of the bowling ball
The momentum of an object is given by the formula: momentum (p) = mass (m) × velocity (v).
The mass of the bowling ball is 7.2 kg, and its initial velocity is 3.5 m/s. Therefore, its initial momentum is:
p_ball_initial = m_ball × v_ball_initial
Step 2: Calculate the initial momentum of the bowling pin
Since the bowling pin is at rest initially, its initial momentum is zero.
p_pin_initial = 0
Step 3: Calculate the total initial momentum
The total initial momentum of the system is the sum of the initial momentum of the bowling ball and the initial momentum of the bowling pin.
total_initial_momentum = p_ball_initial + p_pin_initial
Step 4: Calculate the final momentum of the bowling ball
The final momentum of the bowling ball can be calculated by using the concept of conservation of momentum. According to this principle, the total momentum after the collision is equal to the total momentum before the collision.
p_ball_final = total_initial_momentum
Step 5: Calculate the final momentum of the bowling pin
The final momentum of the bowling pin can also be calculated using the concept of conservation of momentum.
p_pin_final = total_initial_momentum
Step 6: Calculate the magnitude of the final velocity of the bowling ball
The magnitude of the final velocity of the bowling ball can be calculated using the formula for momentum:
p_ball_final = m_ball × v_ball_final
Step 7: Calculate the angle of the final velocity of the bowling ball
To find the angle of the final velocity of the bowling ball, we can use the right triangle formed by the final velocity vector and the original velocity vector. The angle can be found using trigonometry.
Step 8: Determine if the collision is approximately elastic
A collision is considered approximately elastic if the total kinetic energy of the system is conserved. To check if the collision is approximately elastic, we need to calculate the initial and final kinetic energy of the system.
Step 9: Calculate the initial kinetic energy of the system
The kinetic energy of an object is given by the formula: kinetic energy (KE) = 0.5 × mass × (velocity)^2.
Step 10: Calculate the final kinetic energy of the system
Similarly, we can calculate the final kinetic energy of the system.
Step 11: Compare the initial and final kinetic energies
If the initial kinetic energy is approximately equal to the final kinetic energy, then the collision is considered approximately elastic.
I hope this step-by-step guide helps you in solving the problem.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's break it down step by step:
1. Conservation of Momentum:
The initial momentum of the system is equal to the final momentum of the system. We can represent momentum as the product of mass and velocity.
Initial momentum of the system (before the collision) is given by:
P(initial) = m1 * v1 + m2 * v2
where m1 and v1 are the mass and velocity of the bowling ball, and m2 and v2 are the mass and velocity of the bowling pin before the collision.
Final momentum of the system (after the collision) is given by:
P(final) = m1 * v1' + m2 * v2'
where v1' and v2' are the velocities of the bowling ball and pin after the collision.
Since the bowling pin is at rest before the collision, v2 = 0.
The equation for conservation of momentum becomes:
m1 * v1 = m1 * v1' + m2 * v2'
2. Conservation of Kinetic Energy:
For the collision to be approximately elastic, we also need to check if the kinetic energy of the system before and after the collision is conserved.
The initial kinetic energy of the system is given by:
KE(initial) = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
The final kinetic energy of the system is given by:
KE(final) = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Now, let's gather the given information and solve the problem:
m1 (mass of the bowling ball) = 7.2 kg
v1 (velocity of the bowling ball) = 3.5 m/s
m2 (mass of the bowling pin) = 1.2 kg
v2 (velocity of the bowling pin before collision) = 0 m/s
v2' (velocity of the bowling pin after collision) = 3.0 m/s
θ (angle between the final velocity of the bowling pin and the original direction of motion of the bowling ball) = 60°
Using the equations mentioned above, we can calculate the final velocity of the bowling ball:
1. Conservation of Momentum:
m1 * v1 = m1 * v1' + m2 * v2'
(7.2 kg) * (3.5 m/s) = (7.2 kg) * v1' + (1.2 kg) * (3.0 m/s)
25.2 kg*m/s = 7.2 kg * v1' + 3.6 kg*m/s
Simplifying the equation:
21.6 kg*m/s = 7.2 kg * v1'
v1' = 21.6 kg*m/s / 7.2 kg
v1' = 3 m/s
So, the final velocity of the bowling ball is 3.0 m/s.
2. Conservation of Kinetic Energy:
To determine if the collision is approximately elastic, we need to check if the kinetic energy before and after the collision is conserved.
Initial kinetic energy:
KE(initial) = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
KE(initial) = (1/2) * (7.2 kg) * (3.5 m/s)^2 + (1/2) * (1.2 kg) * (0 m/s)^2
KE(initial) = 45.36 J
Final kinetic energy:
KE(final) = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
KE(final) = (1/2) * (7.2 kg) * (3 m/s)^2 + (1/2) * (1.2 kg) * (3.0 m/s)^2
KE(final) = 42.96 J
As KE(initial) is not equal to KE(final), the kinetic energy is not conserved. Therefore, the collision is not approximately elastic.
To summarize, the final velocity of the bowling ball is 3.0 m/s. The collision is not approximately elastic since the kinetic energy is not conserved.