force=change in momentum/time
time=(mass * velocity)-(0)/220 N
t=(0.55 kg*3.2)/220N
=0.008 s
time=(mass * velocity)-(0)/220 N
t=(0.55 kg*3.2)/220N
=0.008 s
The impulse (J) is defined as the product of force (F) and the time interval (Δt) over which the force is applied:
J = F * Δt
The impulse can also be expressed as the change in momentum (Δp) of the ball:
J = Δp
The momentum (p) of an object is defined as the product of its mass (m) and velocity (v):
p = m * v
Therefore, the change in momentum (Δp) can be expressed as:
Δp = m * v - 0
Since the ball is initially at rest, its initial momentum is zero.
Substituting the expression for impulse (J) and change in momentum (Δp) into the equation, we get:
F * Δt = m * v
Rearranging the equation to find the time interval (Δt), we have:
Δt = (m * v) / F
Substituting the given values, we get:
Δt = (0.55 kg * 3.2 m/s) / 220 N
Calculating the value, we find:
Δt = 0.008 s
Therefore, the mallet was in contact with the ball for 0.008 seconds.
Impulse = Change in momentum
Impulse (I) is the product of force (F) and time (Δt) and is given by:
I = F * Δt
Momentum (p) is the product of mass (m) and velocity (v) and is given by:
p = m * v
Since the ball is initially at rest, the initial momentum (pinitial) is zero. After being struck, the ball's final momentum (pfinal) is given by:
pfinal = m * v
The change in momentum (Δp) is then:
Δp = pfinal - pinitial = m * v - 0 = m * v
Using the impulse-momentum theorem, we can equate the impulse with the change in momentum:
I = Δp
F * Δt = m * v
Rearranging the equation to solve for time (Δt):
Δt = (m * v) / F
Now we can substitute the given values into the equation to find the time the mallet was in contact with the ball.
Given:
m (mass of the ball) = 0.55 kg
v (speed of the ball) = 3.2 m/s
F (force exerted on the ball) = 220 N
Substituting the values:
Δt = (0.55 kg * 3.2 m/s) / 220 N
Simplifying:
Δt = 9.36 x 10^-3 s
Therefore, the time the mallet was in contact with the ball is approximately 9.36 milliseconds.