Question

A 140 g baseball is moving horizontally to the right at 35 mis when it is hit by the bat. the ball fl ies off to the le ft at 55 mis, at an angle of 25° above the horizontal. what are the magnitude and direction of the impulse that the bat deli vers to the ball?

Answer 1

Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

When it hits the bat, it flies left with a velocity of 55 m/s at an angle of 25°. To reiterate:

mass = 0.140kg, Initial: 35m/s, Final: 55m/s at an angle of 25°

For this problem, we have to use the impulse equation, but before that lets break the velocity into components (It will be apparent towards the end):

The initial velocity is moving only in the horizontal direction, so:

$v_{0x} = 35 m/s$

The final velocity has an x and a y component:

$v_{fx} = 55cos(25) = 49.84692829m/s\\v_{fy} = 55sin(25) = 23.2440044m/s$

Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

$J = dp= m*dv$

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

$dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s$

Finally, substitute into the equation

$J = m*dv\\J = 0.140kg * 87.97320604m/s\\J = 12.31624885kg*m/s\\J = 12.32 kg*m/s$

J = 12.32kg*m/s