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givi [52]
3 years ago
7

A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface. part a wh

at is the speed of the block after the collision?
Physics
1 answer:
prisoha [69]3 years ago
8 0
Momentum = velocity * mass. Momentum is always conserved. The ball's momentum before the impact is 0.05Vo. After the impact, it is 105Vf. (Vf = final velocity). Because momentum is conserved, we know that:
0.05Vo = 1.05Vf...
Vo = 21Vf...
Vf = Vo / 21.
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Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

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M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

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N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

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\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°

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