A ball of mass 0.81 kg is moving at 4.4 m/s when it smashes into a stationary ba
ID: 1332348 • Letter: A
Question
A ball of mass 0.81 kg is moving at 4.4 m/s when it smashes into a stationary ball with mass 0.62 kg. After the collision the first ball moves at 0.9 m/s, what is the speed of the other ball in m/s? 4.5726
Now the ball of m 0.81 kg moves at 3.2 m/s and hits the second ball (0.62 kg) that moves at 0.6 m/s and after the collision they stick together. What is the final speed of the system in m/s? 2.0727
Now the ball of m 0.81 kg moves at 3.4 m/s and hits the second ball (0.62 kg) that is moving at 3 m/s in opposite direction. After a head collision, the first ball moves backward at 2.4 m/s What is the speed in m/s of the other ball? 4.5774
Now the ball of m 0.81 kg moves at 4.1 m/s and hits a wall bouncing back at 2.8 m/s. The time of contact is 47 ms, What is the magnitude of the force in N exerted on the ball? 118.9149
NOTE: the answers are provided, I just need to know what steps were taken to get these answers and how to calculate them
Explanation / Answer
law of conservation momentum
In the abscence of external force the total momentum of the system remains constant
total momentum before collisions = total momentum after collisions
Pi = Pf
m1*u1 + m2*u2 = m1*v1 + m2*v2
m1 = 0.81 kg m2 = 0.62 kg
u1 = 4.4 m/s u2 = 0
v1 = 0.9 m/s V2 = ?
(0.81*4.4)+(0.62*0) = (0.81*0.9)+(0.62*v2)
v2 = 4.5726 m/s
++++++++++++++++++++
m1 = 0.81 kg m2 = 0.62 kg
u1 = 3.2 m/s u2 = 0.6
v1 = v m/s V2 = v
(0.81*3.2)+(0.62*0.6) = (0.81*v)+(0.62*v)
v = 2.0727 m/s
+++++++++++++++++++++++
m1 = 0.81 kg m2 = 0.62 kg
u1 = 3.4 m/s u2 = -3 m/s
v1 = -2.4 m/s V2 = ?
(0.81*3.4)-(0.62*3) = -(0.81*2.4)+(0.62*v2)
v2 = 4.5774 m/s
++++++++++++++++++++
m1 = 0.81 kg
u1 = 4.1 m/s
v1 = -2.8 m/s
F = m1(V1-u1)/t
F = 0.81*(-2.8-4.1)/0.047 = -118.9149 N
magnitude = 118.9149 N
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