Answer:
A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms, find the horizontal and vertical components of the average force on the ball.
The average horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.
Explanation:
Given:
Mass of the baseball, [tex]m[/tex] = 0.145 kg
Velocity before the impact, [tex]v_1[/tex] = 40 m/s
Velocity after the impact, [tex]v_2[/tex] = 52 m/s
Angle of deflection, [tex]\theta[/tex] = 30 deg
Time of contact, [tex]\triangle t[/tex] = 1.85 ms
Conversion of milliseconds to seconds.
⇒ [tex]\triangle t = 1.85\times 10^-^3\ sec[/tex]
Let the horizontal and vertical component of 'v1' and 'v2' be:
⇒ [tex]v_1x,v_1y,v_2x,v_2y[/tex]
⇒ [tex]v_1x=40\ ms^-^1[/tex] and [tex]v_2y= 0\ ms^-^1[/tex]
⇒ Taking the vector components of 'v2'
⇒ [tex]v_2x=52cos(30) = 45.03\ ms^-^1[/tex] and [tex]v_2y=52sin(30)=40.97\ ms^-^1[/tex]
We have to find the impulse in x-y, direction.
Momentum with 'v1'
⇒ [tex]P_1_x=mv_1_x[/tex]
⇒ [tex]P_1_x=0.145(40)[/tex]
⇒ [tex]P_1_x=0.145(40)[/tex]
⇒ [tex]P_1_x=5.8\ kg.m.s^-^1[/tex] and [tex]P_1_y=0[/tex]
Momentum with 'v2'
⇒ [tex]P_2_x=mv_2_x[/tex] ⇒ [tex]P_2_y=mv_2_y[/tex]
⇒ [tex]P_2_x=0.145(45.03)[/tex] ⇒ [tex]P_2_y=0.145(40.97)[/tex]
⇒ [tex]P_2_x=6.52\ kg.m.s^-^1[/tex] ⇒ [tex]P_2_y=5.96\ kg.m.s^-^1[/tex]
Now the difference in terms of impulse (J).
Horizontally Vertically
⇒ [tex]J_x=P_2_x-P_1_x[/tex] ⇒ [tex]J_y=P_2_y-P_1_y[/tex]
⇒ [tex]J_x=6.52-5.8[/tex] ⇒ [tex]J_y=5.96-0[/tex]
⇒ [tex]J_x=0.72\ kg.m.s^-^1[/tex] ⇒ [tex]J_y=5.96\ kg.m.s^-^1[/tex]
Now we know that Impulse is the product of force and time.
So,
Horizontal force : Vertical force:
⇒ [tex](F_x)_a_v_g=\frac{J_x}{\triangle t}[/tex] ⇒ [tex](F_y)_a_v_g=\frac{J_y}{\triangle t}[/tex]
⇒ [tex](F_x)_a_v_g=\frac{0.72\ kg.ms^-^1}{1.85\times 10^-^3\ s}[/tex] ⇒ [tex](F_y)_a_v_g=\frac{5.96\ kg.m.s^-^1}{1.85\times 10^-^3\ s}[/tex]
⇒ [tex](F_x)_a_v_g=389.18\ N[/tex] ⇒ [tex](F_x)_a_v_g=3221.62\ N[/tex]
So the horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.
