Question 9.55: A 5.0 kg block with a speed of 3.0 m/s collides with a 10 kg......

We choose +x in the direction of (initial) motion of the blocks, which have masses m_1 = 5 kg and m_2 = 10 kg. Where units are not shown in the following, SI units are to be understood.

(a) Momentum conservation leads to

m_{1}\vec{\nu}_{1i}+m_{2}\vec{\nu}_{2i}=m_{1}\vec{\nu}_{1f}+m_{2}\vec{\nu}_{2f}

(5{\mathrm{~kg}})(3.0{\mathrm{~m/s}})+(10{\mathrm{~kg}})(2.0{\mathrm{~m/s}})=(5{\mathrm{~kg}}){\bar{\nu}}_{1f}+(10{\mathrm{~kg}})(2.5{\mathrm{~m/s}})

which yields \vec{\nu}_{1f}=2.0\;\mathrm{m/s}\;. Thus, the speed of the 5.0 kg block immediately after the collision is 2.0 m/s.

(b) We find the reduction in total kinetic energy:

K_{i}-K_{f}=\frac{1}{2}\bigl(5\;\mathrm{kg}\bigr)\bigl(3\;\mathrm{m/s}\bigr)^{2}+\frac{1}{2}\bigl(10\;\mathrm{kg}\bigr)\bigl(2\;\mathrm{m/s}\bigr)^{2}-{\frac{1}{2}}{\Big(}5\mathrm{kg}{\Big)}(2\operatorname{m/s})^{2}-{\frac{1}{2}}{\Big(}10\mathrm{~kg}{\Big)}(2.5\ \mathrm{m/s})^{2}

=-1.25\mathrm{{~J}~}\approx-1.3\mathrm{{~J}}.

(c) In this new scenario where \vec{\nu}_{ 2 f} = 4.0 m/s , momentum conservation leads to \vec{\nu}_{ 1 f} = −1.0 m s and we obtain ΔK = +40 J .

(d) The creation of additional kinetic energy is possible if, say, some gunpowder were on the surface where the impact occurred (initially stored chemical energy would then be contributing to the result).