Respuesta :
Answer:
u = 4.6 m/s
h = 8.01 m
Explanation:
Given:
Mass of the tennis ball, m = 44.0 g
Mass of the basket ball, M = 594 g
Height of fall, h = 1.08m
Now,
we have
[tex]u^2-u'^2 = 2as[/tex]
where, s = distance = h
a = acceleration
u = final speed before the collision
u' = initial speed
since it is free fall case
thus,
a = g = acceleration due to gravity
u' = 0
thus we have
[tex]u^2-0^2 = 2\times9.8\tiimes1.08[/tex]
or
[tex]u = \sqrt{21.168}[/tex]
or
u = 4.6 m/s
b) Now after the bounce, the ball moves with the same velocity
thus, v = v₂
thus,
final speed ([tex]v_f[/tex]) = v = 4.6 m/s
Then conservation of energy says
[tex]\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2[/tex]
also
applying the concept of conservation of momentum
we have
mu₁ + Mu₂ = mv₁ + Mv₂
u₁ =velocity of the tennis ball before collision = -4.6 m/s
u₂ = velocity of the basketball before collision= 4.6 m/s
v₁ = velocity of the tennis ball after collision
v₂ = velocity of the basketball after collision
substituting the values in the equation, we get
Now,
solving both the equations simultaneously we get
[tex]v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2[/tex]
substituting the values in the above equation we get
[tex]v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6[/tex]
or
[tex]v = -8.565-3.965[/tex]
or
[tex]v = -12.53m/s[/tex]
here negative sign depicts the motion of the ball in the upward direction
now the kinetic energy of the tennis ball
[tex]K.E = \frac{1}{2}mv^2[/tex]
or
[tex]K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2[/tex]
or
K.E = 3.45 J
also at the height the K.E will be the potential energy of the tennis ball
thus,
3.45 J = mgh
or
3.45 = 44 × 10⁻³ × 9.8 × h
h = 8.01 m
The height to which the tennis ball rebounds is found with the assumption
that it has equal speed with the basketball before collision.
(a) The magnitude of the downward velocity of the basketball is
approximately 4.6 m/s
(b) The height to which the tennis ball rebounds is approximately 8 meters
Reasons:
Mass of tennis ball, m₁ = 44.0 g
Mass of basketball, m₂ = 594 g
Distance the balls are dropped, h = 1.08 m
(a) The magnitude of the downward velocity, v of the basketball when it
reaches the ground, is given as follows;
v = √(2·g·h)
∴ v = √(2×9.81×1.08) ≈ 4.6
The magnitude of the downward velocity, v ≈ 4.6 m/s
(b) In elastic collision, we have;
[tex]K.E._{itennis}[/tex] + [tex]K.E._{ibasketball}[/tex] = [tex]K.E._{2tennis}[/tex] + [tex]K.E._{2basketball}[/tex]
Which gives;
0.5·m₁·u₁² + 0.5·m₂·u₂² = 0.5·m₁·v₁² + 0.5·m₂·v₂²
Where;
u₂ = -v₂
Therefore;
m₁·u₁ + m₂·u₂ = m₁·v₁ + m₂·v₂
Using the formula for head on elastic collision, we get;
u₁ = The initial velocity of the tennis ball before collision
u₂ = The initial velocity of the basketball before collision
v₁ = The velocity of the tennis ball after collision
v₂ = The velocity of the basketball after collision
[tex]v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]
Which gives;
[tex]v_1 = -\dfrac{25}{29} \cdot u_1 - \dfrac{1242}{145}[/tex]
[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]
Which gives;
[tex]v_2 = \dfrac{4}{29} \cdot u_1 - \dfrac{115}{29}[/tex]
Taking the initial velocity of the tennis ball to be equal and opposite to the
initial velocity of the basketball, we have;
u₁ = 4.6 m/s
Which gives;
[tex]v_1 = -\dfrac{25}{29} \times 4.6- \dfrac{1242}{145} \approx -12.53[/tex]
The initial velocity of the tennis ball upwards, v₁ ≈ 12.53 m/s
The height to which the ball rebounds, [tex]h= \dfrac{v_1^2}{2 \cdot g} = \dfrac{12.53^2}{2 \times 9.81} \approx 8[/tex]
The maximum height to which the tennis ball rebounds, h ≈ 8.0 meters
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