Question

A 166-g hockey puck is gliding across the ice at 19.5 m/s. A player whacks it with her stick, sending it moving at 43.1 m/s at 45.0∘ to its initial direction of motion. What is the angle between the initial velocity of the puck and the force exerted on the puck? If stick and puck are in contact for 112 ms, what is the magnitude of the average force that was exerted on the puck?

Answer 1

Answer:

Explanation:

Given

mass of puck=166 gm

initial velocity=19.5 m/s

Final velocity=43.1 m/s at angle of $45^{\circ}$ to the intial direction

Suppose puck is travelling towards positive x axis

thus initial moment $\vec{P_i}=0.166\times 19.5\hat{i}$

Final momentum $\vec{P_f}=0.166\left ( 43.1cos45\hat{i}+43.1sin45\hat{j}\right )$

Impusle$=P_f-P_i=5.06\hat{i}+5.06\hat{j}-3.237\hat{i}$

$I=1.823\hat{i}+5.06\hat{j}$

Angle between initial velocity and force

$tan\theta =\frac{5.06}{1.823}$

$\theta =70.18^{\circ}$ w.r.t x-axis

and Fdt=Impulse

$dt=112 ms=112\times 10^{-3} s$

$\vec{F}=16.339\hat{i}+45.17\hat{j}$

$|F|=\sqrt{16.34^2+45.17^2}=48.03 N$

Answer 2

The average force is equal to the product of mass and acceleration. The magnitude of the average force that was exerted on the puck will be 48.03 N.

What is the law of conservation of momentum?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The momentum of the body is given by the product of the mass and velocity of the body.

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

The given data in the problem is;

m is the mass of puck=166 gram

u is the initial velocity=19.5 m/s

v is the final velocity=43.1 m/s

The initial momentum of the body is;

$\rm P_i = 0.166 \times 19.5 \vec i$

The final momentum of the body is;

$\rm P_f= 0.166 \times (43.1 cos 45 \vec i + 43.1 sin 45 \vec j)$

The change in the momentum is;

$\rm P_f-\rm P_i = 0.166 \times (43.1 cos 45 \vec i + 43.1 sin 45 \vec j)-0.166 \times 19.5 \vec i \\\\ \rm P= 506.1 \vec i +5.06 \vec j -3.2371 \vec i$

The rate of change of the momentum is the impulse;

$\rm I = \triangle p$

From Newton's second law;

$\rm I = F \triangle t$

$\vec F = 16.339 \vec I +45.71 \vec j \\\\ \vec F = \sqrt{(16.34)^2+(45.147)^2} \\\\ \vec F = 48.03$

Hence the magnitude of the average force that was exerted on the puck will be 48.03 N.

To learn more about the law of conservation of momentum refer to;

https://brainly.com/question/1113396