Question

A hockey puck with mass ﻿0.1600.1600.160﻿ kg is at rest at the origin (﻿x=0x = 0x=0﻿) on the horizontal, frictionless surface of the rink. At time ﻿t=0t = 0t=0﻿ a player applies a force of ﻿0.2500.2500.250﻿ N to the puck, parallel to the ﻿xxx﻿-axis; she continues to apply this force until ﻿t=2.00t = 2.00t=2.00﻿ s. If the same force is again applied at t=5.00t = 5.00 s, what are the position and speed of the puck at t=7.00t = 7.00 s?

Accepted Answer

Step 1: Begin by analyzing the motion of the puck during the first interval (t = 0 to t = 2.00 s). Use Newton's second law, $$ F = ma $$, to calculate the acceleration of the puck. The force $$ F  is $$given as 0.250 N, and the mass $$ m  is 0.160 kg. $$Solve for acceleration $$ a $$ using $$ a = \frac{F}{m} . $$Step 2: Using the calculated acceleration $$ a $$, determine the velocity of the puck at t = 2.00 s. Since the puck starts from rest, the velocity $$ v $$ can be found using the kinematic equation $$ v = v_0 + at $$, where $$ v_0  is $$the initial velocity (0 m/s). Substitute $$ a $$ and $$ t = 2.00  s $$into the equation. Step 3: Calculate the position of the puck at t = 2.00 s using the kinematic equation $$ x = x_0 + v_0t + \frac{1}{2}at^2 . $$Here, $$ x_0  is $$the initial position (0 m), $$ v_0  is $$the initial velocity (0 m/s), and $$ a  is $$the acceleration calculated earlier. Substitute $$ t = 2.00  s $$into the equation. Step 4: For the second interval (t = 5.00 s to t = 7.00 s), note that the puck moves with constant velocity from t = 2.00 s to t = 5.00 s (no force is applied during this time). Use the velocity at t = 2.00 s to calculate the position of the puck at t = 5.00 s using $$ x = x_0 + vt $$, where $$ v  is $$the constant velocity and $$ x_0  is $$the position at t = 2.00 s. Step 5: When the force is reapplied at t = 5.00 s, the puck undergoes acceleration again. Repeat the process from Step 1 to calculate the new acceleration, velocity, and position during the interval t = 5.00 s to t = 7.00 s. Add the displacement during this interval to the position at t = 5.00 s to find the final position at t = 7.00 s. Use $$ v = v_0 + at  to $$find the final velocity at t = 7.00 s.

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주요 개념

Newton's Second Law of Motion

Kinematics Equations

Impulse and Momentum