Question

Use the approximation that Vavg = p m for each time step. A paddle ball toy consists of a flat wooden paddle and a small rubber ball that are attached to each other by an elastic band (figure). You have a paddle ball toy for which the mass of the ball is 0.011 kg, the stiffness of the elastic band is 0.945 N/m, and the relaxed length of the elastic band is 0.280 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 <-0.02,-0.01, -0.02 > kg.m/s, and the moving ball is at location <-0.2,-0.61, 0> m relative to an origin located at the point where the elastic band is attached to the paddle. *Part 1 (a) Determine the position of the ball 0.1 s later, using a At of 0.1 s. (Express your answer in vector form.) 7 = 23m 12 Answer *1: the tolerance is +/-5% Answer *2: the tolerance is +/-5% Answer *3: the tolerance is +/-5% Attempts: 0 of 10 used *Part 2 (b) Starting with the same initial position (<-0.2, -0.61, 0> m) and momentum (<-0.02, -0.01, -0.02 > kg-m/s) determine the position of the ball 0.1 s later, using a At of 0.05 s. (Express your answer in vector form.) 7 =<L Answer *1: the tolerance is +/-5% Answer *2: the tolerance is +/-5% Answer *3: the tolerance is +/-5% Attempts: 0 of 10 used

Answer 1

Initial location $\vec{r}$ = ( -0.2 , -0.61, 0 ) m

normal unit vector $\hat{r}$ = $\vec{r}$ / | $\vec{r}$ | = (-0.311 , -0.950 , 0 ) m

Hence angle made by position vector $\vec{r}$ with respective coordinate axes

cos$\alpha$ = -0.311 , cos$\beta$ = -0.950 , cos$\gamma$ = 0

Elastic force F = k l , where k = 0.945 n/m is stiffness of elastic band and

l is stretched length = |$\vec{r}$| - 0.280 = 0.642 - 0.280 = 0.362 m

magnitude of elastic force = 0.945 $\times$ 0.362 = 0.342 N

Elastic force F = ( F cos$\alpha$ , F cos$\beta$ , 0 ) = ( -0.342 $\times$ 0.311 , -0.342 $\times$ 0.950 , 0) N

Elastic force = (-0.106 , -0.325, 0 ) N

For acceleration, we need to consider elastic force and gravity

Net force = ( -0.106, -0.325, -mg ) N

Resultant acceleration = (-0.106/0.011 , -0.325/0.011 , 9.8) = (-9.636 , -29.545 , -9.8) m/s²

Initial momentum ( -0.02 , -0.01, -0.02 ) kg m/s

Initial velocity( -0.02 / 0.011 , -0.01/0.011 , -0.02 /0.011 ) = (-1.82 , -0.909 , -1.82 ) m/s

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by taking $\Delta$ t=0.1 s

S_x = u_x$\Delta$t + (1/2) a_x$\Delta$t² = -1.82 $\times$ 0.1 +(1/2)(-9.636)$\times$0.1$\times$0.1 = -0.23018 m

S_y = u_y$\Delta$t + (1/2) a_y$\Delta$t² = -0.909 $\times$ 0.1 +(1/2)(-29.545)$\times$0.1$\times$0.1 = -1.056725 m

S_z = u_z$\Delta$t + (1/2) a_z$\Delta$t² = -1.82 $\times$ 0.1 +(1/2)(-9.8)$\times$0.1 $\times$ 0.1 = -0.2065 m

Final position = ( -0.2-0.2302 , -0.61-1.0567, 0-0.2065 ) m = ( -0.4302 , -1.6667, -0.2065 ) m

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by taking $\Delta$ t=0.05 s

for first interval of $\Delta$ t=0.05 s

S_x = u_x$\Delta$t + (1/2) a_x$\Delta$t² = -1.82 $\times$ 0.05 +(1/2)(-9.636)$\times$0.05$\times$0.05 = -0.1030 m

v_x = u_x+a_x$\Delta$t = -1.82 -9.636 $\times$ 0.05 = -1.868 m/s

S_y = u_y$\Delta$t + (1/2) a_y$\Delta$t² = -0.909 $\times$ 0.05 +(1/2)(-29.545)$\times$0.05$\times$0.05 = -0.0824 m

v_y = u_y+a_x$\Delta$t = -0.909 -29.545$\times$0.05 = -2.386 m/s

S_z = u_z$\Delta$t + (1/2) a_z$\Delta$t² = -1.82 $\times$ 0.05 +(1/2)(-9.8)$\times$0.05 $\times$ 0.05 = -0.1033 m

v_z = u_z+a_z$\Delta$t = -1.82 -9.8 $\times$ 0.05 = -2.31 m/s

for next interval of $\Delta$ t=0.05 s

S_x = u_x$\Delta$t + (1/2) a_x$\Delta$t² = -1.868 $\times$ 0.05 +(1/2)(-9.636)$\times$0.05$\times$0.05 = -0.1054 m

S_y = u_y$\Delta$t + (1/2) a_y$\Delta$t² = -2.386 $\times$ 0.05 +(1/2)(-29.545)$\times$0.05$\times$0.05 = -0.1562 m

S_z = u_z$\Delta$t + (1/2) a_z$\Delta$t² = -2.31 $\times$ 0.05 +(1/2)(-9.8)$\times$0.05 $\times$ 0.05 =   -0.1278 m

Net displacement = [ (-0.1030 -0.1054 ) , (-0.0824 -0.1562), ( -0.1033-0.1278 ) ] m

Net displacement = [ -0.2084 , -0.2386 , -0.2311 ] m

Final position = ( -0.2-0.2084 , -0.61-0.2386, 0-0.2311 ) m = ( -0.4084 , -0.8486, -0.2311 ) m