As shown in the figure, a metal ball with mass m2 is initially at rest on a horizontal, frictionless table. A second metal ball with mass m1 moving with a speed 2.00 m/s, collides with m2. Assume m1 moves initially along the +x-axis. After the collision, m1 moves with speed 1.00 m/s at an angle of ? = 50.0° to the positive x-axis.(Assume m1 = 0.200 kg and m2 = 0.300 kg.) (a)Determine the speed (in m/s) of the 0.300 kg ball after the collision.m/s(b)Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.|ΔK| / Ki = ### Collision Analysis: Before and After#### Diagram (a): Before the CollisionThe diagram illustrates two masses, \( m_1 \) and \( m_2 \), approaching each other. Mass \( m_1 \) is moving towards mass \( m_2 \) with an initial velocity \( \vec{v}_{1i} \). This velocity is represented by a red arrow pointing towards \( m_2 \).#### Diagram (b): After the CollisionThe second diagram shows the scenario after the collision between masses \( m_1 \) and \( m_2 \). - **Mass \( m_1 \):** - It travels at a final velocity \( \vec{v}_{1f} \) which is broken down into its horizontal and vertical components: - Horizontal component: \( v_{1f} \cos \theta \) - Vertical component: \( v_{1f} \sin \theta \) - The angle \( \theta \) represents the deviation of \( m_1 \) from its original path.- **Mass \( m_2 \):** - It moves with a velocity \( \vec{v}_{2f} \) which also has horizontal and vertical components: - Horizontal component: \( v_{2f} \cos \phi \) - Vertical component: \( v_{2f} \sin \phi \) - The angle \( \phi \) denotes the direction relative to the original path of \( m_2 \).The diagrams visualize the principles of momentum and kinetic energy conservation in two-dimensional collisions. The angles \( \theta \) and \( \phi \) illustrate the deflection paths post-collision, highlighting the vector nature of velocity components.

Name: As shown in the figure, a metal ball with mass m2 is initially at rest
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: As shown in the figure, a metal ball with mass m2 is initially at rest on a horizontal, frictionless table. A second metal ball with mass m1 moving with a speed 2.00 m/s, collides with m2. Assume m1 moves initially along the +x-axis. After the collis

Given: The masses: m1 = 0.200 kg m2 = 0.300 kg Iintial speed of mass m1 is: v1i=2.00 m/s After the collision, the speed of mass m1 is: v1f=1.00 m/s The angle after the collision: θ=50° First, let us apply the conservation of momentum along the x-axis: m1v1i = m1v1fcosθ + m2v2fcos Φor, 0.2×2=0.2×1×cos50°+ 0.3×v2fcos Φor, 0.3×v2fcos Φ=0.4-0.13or, 0.3×v2fcos Φ=0.27or, v2fcos Φ=0.9 ----(1) Now, from the conservation of momentum along the y-axis: m1v1fsinθ - m2v2fsin Φ=0or, m2v2fsin Φ=m1v1fsinθor, 0.3×v2fsin Φ=0.2×1×sin°or, 0.3×v2fsin Φ=0.153or, v2fsin Φ=0.51 -----(2) Now, let us square and add equations 1 and 2. v2fcos Φ2+v2fsin Φ=0.9 2+0.512or, v2f2=0.81+0.26or, v2f=1.03 m/s The speed of the 0.300 kg ball after the collision is 1.03 m/s. To find the direction, let us divide 2 by 1. v2fsin Φv2fcos Φ=0.51 0.9or, tanΦ=0.57or, Φ=29.54° This angle is below the positive x-axis as shown in the given figure.Let us first calculate the initial kinetic energy: Ki=12m1vi2+0or, Ki=12×0.2×2.002or, Ki=0.4 J The final Kinetic energy of the system will be: Kf=12m1v1f2+12m2v2f2or, Kf=12×0.2×1.002+12×0.3×1.032or, Kf=0.26 J Therefore, the change in kinetic energy: ∆K=Ei-Ek=0.4-0.26=0.14 J ∴∆KKi=0.140.4=0.35 The fraction of kinetic energy transferred away or transformed to other forms of energy in the collision is 0.35.