Physics - Elastic Collisions in 2 or 3 Dimensions

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Elastic Collisions in 2 or 3 Dimensions

Table of Contents

Introduction

Problems

Answers

Introduction

When the collision involves 2 or 3 dimensions, the vector nature of momentum must be taken into account. Kinetic Energy is not a vector.

The deflection angle is measured from the initial motion of the first particle.

The Conservation of Energy creates one equation.

The Conservation of Momentum creates one equation for every dimension.

The number of independent equations must be equal to the number of unknowns. This means that we must know some of the conditions after the collision.

Charged particles may have deflections before the collision takes place. We will ignore this in the problems for this section.

Table of Contents

Problems

1.	A neutron, 1.009 amu is shot at a stationary helium-4 atom, 4.01 amu with a speed of 4.55 x 10⁴ m s^-1. The neutron is deflected at an angle 25.5^o to the left of its original path at 3.00 x 10⁴ m s^-1. What is the velocity of the helium-4 atom?
2.	While playing marbles, a shooter with twice the mass of a marble travels at 0.250 m/s until it hits a marble. If the shooter deflects at 35.0^o to the left and the marble leaves at 50.0^o to the right, what is the speed of the marble and the shooter?
3.	A 115 kg guard running forward at 7.00 m/s collides with a 75.0 kg wide receiver running 9.00 m/s coming from the right and perpendicular to the path of the guard. If the guard bounces off at 5.00 m/s and 40.0^o to the left of his original path, what is the velocity of the wide receiver after the collision?
4.	Use conservation of momentum to show that in a two particle collision, the paths of the particles before and after the collision all lie in one plane if (a) one particle is initially at rest; (b) the two particles have momentum in the same or opposite direction?

Table of Contents

Answers

1.	A neutron, 1.009 amu is shot at a stationary helium-4 atom, 4.01 amu with a speed of 4.55 x 10⁴ m s^-1. The neutron is deflected at an angle 25.5^o to the left of its original path at 3.00 x 10⁴ m s^-1. What is the velocity of the helium-4 atom?
	Conservation of Energy v_n = v'_He - v'_n Conservation of Momentum - X m_nv_n = m_nv'_nCos_n + m_Hev'_HeCos_He Conservation of Momentum - Y 0 = m_nv'_nSin_n - m_Hev'_HeSin_He We have three independent equations and two unknowns. Solve Conservation of Energy for v'_He. v'_He = v'_n + v_n = 3.00 x 10⁴ m s^-1 + 4.55 x 10⁴ m s^-1 = 7.55 x 10⁴ m s^-1 Use Conservation of Momentum - X to solve for _He. m_nv_n - m_nv'_nCos_n _He = Cos^-1 ( ------------------------------ ) m_Hev'_He 1.009 amu x 4.55 x 10⁴ m s^-1 - 1.009 amu x 3.00 x 10⁴ m s^-1 Cos 25.5^o _He = Cos^-1 ( -------------------------------------------------------------------------------------------------- ) = 86.5^o 4.01 amu x 7.55 x 10⁴ m s^-1

2.	While playing marbles, a shooter with twice the mass of a marble travels at 0.250 m/s until it hits a marble. If the shooter deflects at 35.0^o to the left and the marble leaves at 50.0^o to the right, what is the speed of the marble and the shooter?
	Conservation of Energy v_s = v'_m - v'_s Conservation of Momentum - X m_sv_s = m_sv'_sCos_s + m_mv'_mCos_m Conservation of Momentum - Y 0 = m_sv'_sSin_s - m_mv'_mSin_m m_s = 2m_m Substitute values for Sines and Cosines into Momentum Equations. Also substitute 2m_m for m_s. 2m_mv_s = 2m_mv'_sCos 35.0^o + m_mv'_mCos 50.0^o 0 = 2m_mv'_sSin 35.0^o - m_mv'_mSin 50.0^o 2m_mv_s = 1.64 m_mv'_s + 0.643 m_m v'_m 0 = 1.15 m_mv'_s - 0.766 m_m v'_m Add the two equations. 2m_mv_s = 1.64 m_mv'_s + 1.15 m_mv'_s + 0.643 m_m v'_m - 0.766 m_m v'_m 2m_mv_s = 2.79 m_mv'_s - 0.123 m_m v'_m Divide both sides by 2m_m. v_s = 1.395 v'_s - 0.0615 v'_m Solve Conservation of Energy for v'_m and Substitute into the previous equation. v_s = v'_m - v'_s => v'_m = v_s + v'_s v_s = 1.395 v'_s - 0.0615 v_s - 0.0615 v'_s Solve for v'_s. ( 1.395 - 0.0615 ) v'_s = ( 1.0000 - 0.7045 ) v_s ( 1.0000 - 0.7045 ) 0.250 m/s v'_s = --------------------------------------- = 0.0554 m/s ( 1.395 - 0.0615 ) Solve the Conservation of Energy for v'_m. v_s = v'_m - v'_s => v'_m = v_s + v'_s = 0.250 m/s + 0.0554 m/s = 0.305 m/s

3.	A 115 kg guard running forward at 7.00 m/s collides with a 75.0 kg wide receiver running 9.00 m/s coming from the right and perpendicular to the path of the guard. If the guard bounces off at 6.00 m/s and 40.0^o to the left of his original path, what is the velocity of the wide receiver after the collision?
	Conservation of Energy v_g - v_w= v'_w - v'_g Conservation of Momentum - X m_wv_w = m_gv'_gCos_g - m_wv'_wCos_w Conservation of Momentum - Y m_gv_g = m_gv'_gSin_g + m_wv'_wSin_w Solve Conservation of Energy for v'_w. v'_w = v_g + v_w+ v'_g = 7.00 m/s - 9.00 m/s + 6.00 m/s = 4.00 m/s Solve Conservation of Momentum - X for _w. m_gv'_gCos_g - m_wv_w _w = Cos^-1 ( ------------------------------ ) m_wv'_w 115 kg x 6.00 m/s x Cos 40.0^o - 75.0 kg x 9.00 m/s _w = Cos^-1 ( ---------------------------------------------------------------------- ) = 119^o 75.0 kg x 4.00 m/s This is not a logical answer because the wide receiver would be bounced forward. _w must be on the same side as _g. Conservation of Momentum - X becomes m_wv_w = m_gv'_gCos_g + m_wv'_wCos_w Now when we solve Conservation of Momentum - X for _w. 75.0 kg x 9.00 m/s - 115 kg x 6.00 m/s x Cos 40.0^o _w = Cos^-1 ( ---------------------------------------------------------------------- ) = 60.8^o 75.0 kg x 4.00 m/s

4.	Use conservation of momentum to show that in a two particle collision, the paths of the particles before and after the collision all lie in one plane if (a) one particle is initially at rest; (b) the two particles have momentum in the same or opposite direction?

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