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Q105P

Expert-verifiedFound in: Page 255

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A 3.0 kg object moving at 8.0 m/s** ** in the positive direction of an x axis has a one-dimensional elastic collision with an object of mass M, initially at rest. After the collision the object of mass M has a velocity of **

Mass of the object, M = 5.0 kg .

- Mass, m = 3.0 kg .
- The velocity of mass m, ${v}_{i1}=8.0\mathrm{m}/\mathrm{s}$ .
- The velocity of mass M, ${v}_{f2}=6.0\mathrm{m}/\mathrm{s}$ .

**You use the concept of conservation of momentum and conservation of kinetic energy. You can write the two equations, one for conservation of momentum and another for conservation of kinetic energy. You will have two unknowns; you can rearrange one equation for one of the unknowns, and by plugging it into another equation, you will get an equation with only one unknown, and then you can solve for that unknown. **

We know the initial velocity of mass M is ${v}_{i2}=0$, and the final velocity of mass m ${v}_{f1}$ is unknown.

From the conservation of momentum principle, we can write,

${m}_{1}{v}_{i1}+{m}_{2}{v}_{i2}={m}_{1}{v}_{f1}+{m}_{2}{v}_{f2}$

We plug the values in the equation of conservation of momentum, and we get,

$\left(3.0\right)\left(8.0\right)+M\left(0\right)=\left(3.0\right){v}_{f1}+M(6.0)\phantom{\rule{0ex}{0ex}}24=\left(3.0\right){v}_{f1}+\left(6.0\right)M$

We can rearrange this equation for , and we get,

$24-\left(6.0\right)M=\left(3.0\right){v}_{f1}\phantom{\rule{0ex}{0ex}}{v}_{f1}=8-2M$ (1)

From the kinetic energy principle, we can write,

$\left(\frac{1}{2}\right){m}_{1}{\left({v}_{i1}\right)}^{2}+\left(\frac{1}{2}\right){m}_{2}{\left({v}_{i2}\right)}^{2}=\left(\frac{1}{2}\right){m}_{1}{\left({v}_{f1}\right)}^{2}+\left(\frac{1}{2}\right){m}_{2}{\left({v}_{f2}\right)}^{2}$

Plugging the values in the equation of conservation of kinetic energy, we get,

$\left(\frac{1}{2}\right)\left(3.0\right){\left(8.0\right)}^{2}+\left(\frac{1}{2}\right)M{\left(0\right)}^{2}=\left(\frac{1}{2}\right)\left(3.0\right){\left({v}_{f1}\right)}^{2}+\left(\frac{1}{2}\right)M{\left(6.0\right)}^{2}$

We can cancel $\left(\frac{1}{2}\right)$ from both sides because it is common on both sides; we get,

$\left(3.0\right){\left(8.0\right)}^{2}=\left(3.0\right){\left({v}_{f1}\right)}^{2}+M{\left(6.0\right)}^{2}\phantom{\rule{0ex}{0ex}}192=\left(3.0\right)){\left({v}_{f1}\right)}^{2}+36M$ (2(

Using the value of from equation (1) in equation (2), we get,

$192=\left(3.0\right)){\left(8-2M\right)}^{2}+36M\phantom{\rule{0ex}{0ex}}192=\left(3.0\right)\left(64-32M+4{M}^{2}\right)+36M\phantom{\rule{0ex}{0ex}}192=192-96M+12{M}^{2}+36M$

Solving further as,

$12{M}^{2}-60M=0\phantom{\rule{0ex}{0ex}}12M=60M\phantom{\rule{0ex}{0ex}}M=\frac{60}{12}\phantom{\rule{0ex}{0ex}}=5.0\mathrm{kg}$

Thus, the mass of the object is, M = 5.0 kg .

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