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3 years ago

15

A 0.17 kg hockey puck slows down trom 54 m/s to 35 m/s when it slides on horizontal ice surface. Find the change in momentum of

the puck?

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

D_momentum = 3.23[kg*m/s]

Explanation:

The change of moment can be easily determined by applying the principle of momentum and amount of movement

initial data are:

$m_{1}=0.17[kg]\\ m_{2}=0.17[kg]\\v_{1}=54[m/s]\\v_{2}=35[m/s]\\D_{momentum}= m_{1}*v_{1}-m_{2}*v_{2}\\D_{momentum}=0.17(54-35)\\D_{momentum}= 3.23[kg*m/s]$

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Answer:

a. $I=981.34 N*s$

b. $v_f=3.96 m/s$

c. $v_{f1}=3.63m/s$

d. $y_f=0.673m$

Explanation:

Given: $m=67kg$ , $h=0.720m$ , $0$

a.

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

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$I=981.34 N*s$

b.

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

c.

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

d.

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$

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The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7

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Answer:

$\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=3.32344\times 10^{-7}\ m$

Explanation:

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c = Speed of light = $3\times 10^8\ m/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

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$v_B$ = Velocity of B particle = $5.1\times 10^5\ m/s$

The wavelength is given by

$\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}$

$\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m$

The wavelength $\lambda_A=2.65177\times 10^{-7}\ m$

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The wavelength $\lambda_B=3.32344\times 10^{-7}\ m$

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