Calculate Impulse and average force

[Starrrrr]

1. A baseball of mass 0.1 kg is moving horizontally at a speed of 40 m/s when it is stuck by a bat. It leaves the bat in a direction at an angle φ = 30◦ above its incident path and with a speed of 50 m/s.

• Find the impulse the bat exerts on the ball.
• Assuming the collision lasts for 0.0015 s, what is the average

force the bat is exerting on the ball during the impact?

2. F=m(v-u)/t and I=f x change in temperature


3. My attempt was: I subbed the figures into the force equation like so f=0.1(50-40)/0.0015=666.6
I know this is incorrect because I don't know what to do with the angle that was given in the question.

 

[TJGilb]

Keep in mind that velocity is a vector. It has both direction, and magnitude. You've accounted for the magnitude, but not the direction. Let's say we're in Cartesian coordinates, and that the ball is initially traveling along the positive x-axis at 40 m/s. As a vector, we could say this is <40,0,0>. What will the velocity vector be as it's leaving the bat?

 

[Starrrrr]

Keep in mind that velocity is a vector. It has both direction, and magnitude. You've accounted for the magnitude, but not the direction. Let's say we're in Cartesian coordinates, and that the ball is initially traveling along the positive x-axis at 40 m/s. As a vector, we could say this is <40,0,0>. What will the velocity vector be as it's leaving the bat?

I thing I figured out the impulse so what I got was (8.33i+2.5j) kg m/s

 

[TJGilb]

What does your x represent in your equation for impulse?

 

[Starrrrr]

What does your x represent in your equation for impulse?

Multiply. I got the magnitude of the force to be 5797.99 N

 

[haruspex]

Multiply. I got the magnitude of the force to be 5797.99 N

Looks good, but you should not specify so many significant figures. Round it to two.

 

[TJGilb]

Show me how you plugged in your calculations to find that.

 

[Starrrrr]

Show me how you plugged in your calculations to find that.

Impulse:
momentum of the ball prior to impact is p(o)=mv(o)=o
p(o)=(0.1)(-40i)=-4i kg m/s

momentum of the ball after impact is
p(t)=mv(t)
0.1(50cos(30i)+50sin(30)j)= (4.33i+2.5j) kg m/s
I=p(t)-p(o)=(8.33i+2.5j) kg m/s

Average force: Fav=1/t(I)=1/0.0015((8.33i+2.5j)= 5553.3i+1666.6j N
Magnitude of force is |Fav|= sqrt(5553.3^2+1666.6^2)= 5797.99 N

 

[TJGilb]

Nvm, that looks right. I must have plugged it wrong into my calc (missed a negative). But like haruspex said that's more sig figs than you need.

 

[Starrrrr]

Nvm, that looks right. I must have plugged it wrong into my calc (missed a negative). But like haruspex said that's more sig figs than you need.

Ok, thanks for the help! :)

 

[Starrrrr]

Looks good, but you should not specify so many significant figures. Round it to two.

Thanks! :)