Answer:
Velocity of the ping pong ball must be = V2= 6,035.34m/s
Explanation:
M1= momentum of the bowling ball
m1 = mass of the bowling ball= 5.8kg
v1= velocity of the bowling ball= 1.59m/s
M2= momentum of the ping pong ball
m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg
v2= velocity of the ping pong ball
Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s
Momentum of the ping pong ball = M2= M1= m2v2
= 0.001528 *v2= 9.222
v2= 9.222/0.001528= 6,035.34 m/s
It will decrease
When the temperature increased, the rate will decrease
The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
Learn more about Conservation of Linear Momentum here:
brainly.com/question/7538238
Answer:
i d k about that but I know it`s a Polish thing
