Question

A hockey player makes a slap shot exerting a constant force of 40.0 newtons on the puck for .2 seconds. What is the magnitude of the impulse given to the puck?

Answer 1

The magnitude of the impulse given to the puck by the hockey player's slap shot is 8.0 Ns, calculated by multiplying the constant force of 40.0 newtons by the duration of 0.2 seconds.

Step-by-step explanation:

The magnitude of the impulse given to the puck can be calculated using the definition of impulse, which is the product of the force applied and the time for which it is applied. Impulse (J) is given by the equation J = F × t, where F is the force in newtons and t is the time in seconds.

For the hockey player's slap shot:

Force applied, F = 40.0 newtons

Time of contact, t = 0.2 seconds

Using the formula:

J = 40.0 N × 0.2 s = 8.0 Ns (Newton-seconds)

The magnitude of the impulse given to the puck is 8.0 Ns.

Answer 2

Magnitude of impulsive force is 8 N.s.

Step-by-step explanation:

Given:

Force acting on the puck is,
`F=40.0\ N`

Time interval for which the force acts is,
`\Delta t=0.2\ s`

We are asked to find the impulsive force.

Impulsive force acting on a body is the sudden change in the momentum of the body due to the application of a large constant force for a very small interval of time.

Here, the hockey player applies a large force of 40 N for a very short interval of time. So, the impact on the puck is measured as Impulse and is represented by 'J'.

Impulse in terms of applied force and time interval is given as:

`J=F\Delta t`

Plug in the given values and solve for 'J'. This gives,

`J=40* 0.2\\J=8\ N\cdot s`

Therefore, the magnitude of impulsive force is 8 N.s.