a hockey puck of mass 0.05kg sliding on the ice with velocity 20m/s is stopped by a goalie in 0.10sec. what is the magnitude and direction of the force exerted by the goalie on the puck.
Impulse=change momentum
force*time=mass*(0-20)
force= .05*-20/.1 notice the direction is opposite the initial velocity
while blocking shots is great, its also means the play is in our end....hold on to the puck, cratee more offense, and we will play in our own end less therefore have less shots to block
To find the magnitude and direction of the force exerted by the goalie on the puck, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma).
First, let's calculate the acceleration of the hockey puck. We can use the formula for acceleration, which is the change in velocity divided by the time taken.
Acceleration (a) = (final velocity - initial velocity) / time taken
a = (0 - 20 m/s) / 0.10 s
a = -200 m/s²
Since the velocity changes from 20 m/s to 0 m/s, we have a negative acceleration value.
Next, we can find the force exerted by the goalie on the puck using Newton's second law.
Force (F) = mass (m) * acceleration (a)
F = 0.05 kg * (-200 m/s²)
F = -10 N
The negative sign indicates that the force exerted by the goalie is opposite in direction to the motion of the puck.
Therefore, the magnitude of the force exerted by the goalie on the puck is 10 N, and the direction is opposite to the motion of the puck.