Q44P A paddle ball toy consists of a ... [FREE SOLUTION]

Chapter 2: Q44P (page 86)

A paddle ball toy consists of a flat wooden paddle and asmall rubber ball that are attached to each other by an elastic band (Figure 2.61). You have a paddle ball toy for which the mass of the ball is 0.015 kg, the stiffness of the elastic band is 0.9 N/m, and the relaxed length of the elastic band is 0.30 m. You are holding the paddle so the ball hangs suspended under it, when your cat comes along and bats the ball around, setting it in motion. At a particular instant the momentum of the ball is $鉄� - 0.02, - 0.01, - 0.02 鉄� kg 鈰� m / s$ , and the moving ball is at location $鉄� - 0.2, - 0.61, 0 鉄� m$ relative to an origin located at the point where the elastic band is attached to the paddle.
(a) Determine the position of the ball 0.1 s later, $螖 t$ using of 0.1 s.
(b) Starting with the same initial position
$(鉄� - 0.2, - 0.61, 0 鉄� m)$ and momentum $(鉄� - 0.02, - 0.01, - 0.02 鉄� kg 鈰� m / s)$ , determine the position of the ball 0.1 s later, using a of 0.05 s. (c) If your answers are different, explain why.

Short Answer

Expert verified

(a) The position of the ball 0.1 s later, using $鈭� t$ of 0.1 s, is $鉄� - 0.301, - 0.629, - 0.133 鉄� m$

(b) The position of the ball 0.1 s later, using of 0.5 s, is $r_{f} = 鉄� - 0.259, - 0.631, - 0.067 鉄� m .$ .

(c) Yes, the answer is different. This is because of using the different time steps. When we use smaller time steps, we will get a more accurate result.

Step by step solution

Identification of given data

Mass of the ball is, $m = 0.15 kg$

Stiffness of elastic band, $k_{s} = 0.9 N / m$

Relaxed strength, $L_{0} = 0.30 m$

The initial Moment of inertia of the ball $(\overset{鈫�}{p_{i}}) = 鉄� - 0.02, - 0.01, - 0.02 鉄� kg 鈰� m / s$

The initial location of the ball is given $r_{i} = 鉄� - 0.2, - 0.61, 0 鉄� m$

Formula used to solve the problem.

Gravitational force

$\overset{鈫�}{F} = \overset{鈫�}{m g}$

Force by a spring

$\overset{鈫�}{F} = - k_{s} s \hat{L}$

The s is stretched, and the vector $\hat{L}$ extends from the point of attachment of the

spring to the movable end. $k_{s}$ is the 鈥渟pring stiffness鈥� (also called 鈥渟pring constant鈥�).

(a) Determine the position of the ball 0.1 s later, using ∆t of 0.1 s.

The forces acting on the ball are gravitational force and the force of spring

Gravitational force can be given as

$\overset{鈫�}{F} = \overset{鈫�}{m g} = (0.015 k g) (<0, 9.8, 0> m / s^{2}) = <0, - 0.147, 0> N$

Spring forces

$F_{s p r i n g} = - k_{s} s \hat{L} = - (0.9 N / m) (0.30 m) <0, 31, - 0.95, 0> = <0.31, - 0.95, 0> N = <0.095, 0.291, 0> N$

$\overset{鈫�}{F_{n e t}} = \overset{鈫�}{F_{g}} + {\overset{鈫�}{F}}_{s p r i n g} = <0, - 0.147, 0> N + <0.095, 0.291, 0> N = <0.095, 1.44, 0> N$

We know that the position of the ball can be described by the equation

$r_{f} = r_{1} + \frac{\overset{鈫�}{p_{i}}}{m} 螖 t + \frac{1}{2} \frac{\overset{鈫�}{F_{n e t}}}{m} 螖 t^{2}$

Where $r_{f}$ final position, $r_{j}$ initial position, the $p_{i}$ initial momentum $\overset{鈫�}{F_{n e t}}$ is net force and

$鈭� t$ is time step.

Now substituting the values into the equation

$r_{f} = 鉄� - 0.2, - 0.61, 0 鉄� m + \frac{(鉄� - 0.02, - 0.01, - 0.02 鉄� kg 鈰� m / s)}{(0.015 kg)} (0.1 s) + \frac{1}{2} \frac{鉄� 0.095, 1.44, 0 鉄� N}{(0.015 kg)} (0.1 s)^{2} r_{f} = 鉄� - 0.2, - 0.61, 0 鉄� m + 鉄� - 0.133, - 0.067, - 0.133 鉄� m + 鉄� 0.032, 0.048, 0 鉄� m r_{f} = 鉄� - 0.301, - 0.629, - 0.133 鉄� m$

The position of the ball 0.1 s later, using $鈭� t$ of 0.1 s, is $鉄� - 0.301, - 0.629, - 0.133 鉄� m$

(b) Determine the position of the ball 0.1 s later, using ∆t of 0.5 s.

From part (a)

We know,

${\overset{鈫�}{F}}_{n e t} = 鉄� 0.095, 1.44, 0 鉄� N$

Again, using the equation

$r_{f} = r_{1} + \frac{\overset{鈫�}{p_{i}}}{m} 螖迟 + \frac{1}{2} \frac{F_{net}}{m} {螖迟}^{2}$

Where $r_{f}$ final position, $r_{i}$ initial position, the $p_{j}$ initial momentum $\overset{鈫�}{F_{n e t}}$ is net force and

$鈭� t$ is time interval

Now substituting the values into the equation

$r_{f} = 鉄� - 0.2, - 0.61, 0 鉄� m + \frac{(鉄� - 0.02, - 0.01, - 0.02 鉄� kg 鈰� m / s)}{(0.015 kg)} (0.5 s) + \frac{1}{2} \frac{<0.095, 1.44, 0> N}{(0.015 kg)} (0.5 s)^{2} a r_{f} = 鉄� - 0.2, - 0.61, 0 鉄� m + 鉄� - 0.67, - 0.033, - 0.067 鉄� m + 鉄� 0.008, 0.012, 0 鉄� m r_{f} = 鉄� - 0.259, - 0.631, - 0.067 鉄� m$

The position of the ball 0.1 s later, using $鈭� t$ of 0.5 s, is $r_{f} = 鉄� - 0.259, - 0.631, - 0.067 鉄� m$

(c) Explanations

Yes, in both the cases we are getting the different answers, with of 0.1 s the final position is $鉄� - 0.301, - 0.629, - 0.133 鉄� m .$ and for $鈭� t$ of 0.5 s, the final position is $r_{f} = 鉄� - 0.259, - 0.631, - 0.067 鉄� m .$ This is because of using the different time steps. When we use smaller time steps, we will get a more accurate result.