A 0.70 kg ball is moving horizontally with a speed of 8.8 m/s when it strikes a vertical wall. The ball rebounds with a speed of 3.7 m/s. What is the magnitude of the change in linear momentum of the ball?
-I am not sure how to honestly to figure this one out.
Mass times velcity change.
0.70 kg*[8.8 -(-3.7] m/s
= 0.70 kg*12.4 m/s
= ____ kg*m/s
To find the magnitude of the change in linear momentum of the ball, we need to calculate the initial and final linear momenta and then find the difference between them.
1. Linear momentum (momentum) is defined as the product of an object's mass and velocity, given by the equation:
momentum = mass * velocity
2. The initial linear momentum (p1) of the ball is:
p1 = mass * initial velocity
Substituting the given values:
mass = 0.70 kg
initial velocity = 8.8 m/s
p1 = 0.70 kg * 8.8 m/s
3. The final linear momentum (p2) of the ball is:
p2 = mass * final velocity
Substituting the given values:
mass = 0.70 kg
final velocity = 3.7 m/s
p2 = 0.70 kg * 3.7 m/s
4. Now, find the magnitude of the change in linear momentum (Δp):
Δp = |p2 - p1|
Substituting the calculated values of p2 and p1:
Δp = |0.70 kg * 3.7 m/s - 0.70 kg * 8.8 m/s|
Δp = |0.70 kg * (3.7 m/s - 8.8 m/s)|
Δp = |0.70 kg * (-5.1 m/s)|
Δp = 0.70 kg * 5.1 m/s
5. Now, multiply the mass and the velocity to find the magnitude of the change in linear momentum:
Δp = 3.57 kg·m/s
Therefore, the magnitude of the change in linear momentum of the ball is 3.57 kg·m/s.