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Assoli18 [71]
3 years ago
12

Blocks A (5.00 kg) and B (10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is mov

ing toward it at 4.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. What is the maximum energy stored in the spring bumpers during the collision process
Physics
1 answer:
salantis [7]3 years ago
7 0

Answer: 1.33m/s: 13.326j

Explanation:

M1U1+M2U2=V(M1+M2)

M1=5kg, M2=10kg

U1=4.00m/s, U2=0

5*4+10*0=V(10+5)

20+0=15V

20=15V

V=20/15

V=1.333m/s

K. E=1/2MV^2

K. E= 1/2*15*1.332^2

K. E= 1/2*15*1.77689

K. E=26.65335/2

K. E=13.326j

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3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

\omega_f^2 - \omega_i ^2 =2 \alpha \theta

where

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Re-arranging the formula, we can find \alpha:

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2

Now we want to know the time the bit takes starting from rest to reach a speed of \omega_f=7.85\cdot 10^4 rad/s. So, we can use the following equation:

\alpha = \frac{\omega_f-\omega_i}{t}

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t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s

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