Question

A batter hits a foul ball. The 0.14-kg baseball that was approaching him at 40 m/s leaves the bat at 30 m/s in a direction perpendicular to the line between the batter and the pitcher. What is the magnitude of the impulse delivered to the baseball

Answer 1

Answer:

The magnitude of the impulse delivered to the baseball is 9.8 N s

Explanation:

Given data:

m = mass = 0.14 kg

vi = initial velocity = 40 m/s

vf = final velocity = -30 m/s

The magnitude of the impulse is equal to:

$J=m(v_{i} -v_{f} )=0.14*(40-(-30))=9.8Ns$

Answer 2

Answer:

7 kgm/s or 7 Ns

Explanation:

Since the velocity are perpendicular to each other, the impulses will also be perpendicular.

From newton's second law of motion,

Impulse = mass×velocity

I = mv............. Equation 1

For the horizontal,

I₁ = mv₁......... Equation 2

Given: m = 0.14 kg, v₁ = 40 m/s

Substitute into equation 2

I₁ = 40(0.14)

I₁ = 5.6 kgm/s

Also for the vertical,

I₂ = mv₂.............. Equation 3

Given: m= 0.14 kg, v₂ = 30 m/s

Substitute into equation 3

I₂ = 0.14(30)

I₂ = 4.2 kgm/s

Using Pythagoras to get the resultant impulse

I = √(I₁²+I₂²)................ Equation 4

Substitute the value of I₁ and I₂ into equation 4

I = √(5.6²+4.2²)

I = √(31.36+17.64)

I = √49

I = 7 kgm/s or 7 Ns