{"id":2308,"date":"2018-04-30T18:19:00","date_gmt":"2018-04-30T10:19:00","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2308"},"modified":"2023-08-08T12:45:13","modified_gmt":"2023-08-08T12:45:13","slug":"momentum-impulse-and-projectile-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/momentum-impulse-and-projectile-motion-problems-and-solutions.htm","title":{"rendered":"Momentum Impulse and Projectile motion – Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Momentum Impulse and Projectile motion – Problems and Solutions<\/strong><\/span><\/p>\n

1. A 0.2-kg ball will be inserted into hole C, as shown in the figure below. Hitter strikes the ball in 0.01 second and the path of B-C traveled in 1 second. Determine the magnitude of the force so the ball can be inserted into hole C. Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>.<\/span><\/p>\n

Known :\"Momentum,<\/u><\/span><\/p>\n

Angle (\u03b8) = 60o<\/sup><\/span><\/p>\n

Mass<\/a> of ball (m) = 0.2 kg<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Time interval (\u0394t) = 0.01 second<\/span><\/p>\n

Time interval to travel path B-C (t) = 1 second<\/span><\/p>\n

Wanted :<\/u> Force (F)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Equation of impulse<\/a> : I = F \u0394t <\/span><\/p>\n

Equation of the change in momentum<\/a> : \u0394p = m (vt <\/sub>\u2013 vo<\/sub>).<\/span><\/p>\n

The impulse equals the change in momentum<\/a> :<\/span><\/p>\n

I = \u0394p<\/span><\/p>\n

F \u0394t = m (vt <\/sub>\u2013 vo<\/sub>)<\/span><\/p>\n

F = m (vt<\/sub> \u2013 vo<\/sub>) \/ \u0394t<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

\u0394t = 0.01 second<\/span><\/p>\n

m = 0.2 kg<\/span><\/p>\n

v<\/i>t <\/i><\/sub>= <\/i>the final speed in the equation of impulse-momentum <\/i>= <\/i>the initial speed of ball <\/i>(v<\/i>o<\/i><\/sub>) <\/i>in the projectile motion<\/i><\/span><\/p>\n

vo<\/sub> = the initial speed in the equation of impulse-momentum = 0 m\/s (initially ball at rest)<\/span><\/p>\n

F = m (vt <\/sub>\u2013 vo<\/sub>) \/ \u0394t<\/span><\/p>\n

F = 0.2 (vt <\/sub>\u2013 0) \/ 0.01<\/span><\/p>\n

F = 0.2 vt <\/sub> \/ 0,01<\/span><\/p>\n

continued<\/span><\/span>……<\/em><\/span><\/p>\n

Determine the initial speed of the ball <\/b><\/u>(v<\/b><\/u>o<\/b><\/u><\/sub>) <\/b><\/u>in projectile motion<\/b><\/u><\/span><\/p>\n

Since the ball is hit until the ball reaches the point B = part 1 of the projectile motion<\/a>.<\/span><\/p>\n

The ball travels from point B to C = part 2 of the projectile motion.<\/span><\/p>\n

Part 2 of the projectile motion :<\/b><\/span><\/p>\n

The projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at a constant velocity and the y motion occurs at a constant acceleration of gravity. <\/span><\/p>\n

\"Momentum,Known :<\/u><\/span><\/p>\n

Horizontal distance (x) = 5 meters<\/span><\/p>\n

Time in air (t) = 1 second<\/span><\/p>\n

x and t are known so that vox<\/sub> can calculated using the equation of the uniform linear motion<\/a>. vox<\/sub> is the horizontal component of initial speed of ball.<\/span><\/p>\n

vox<\/sub> = x \/ t = 5 meters \/ 1 second = 5 m\/s.<\/span><\/p>\n

Part 1 of the projectile motion :<\/b><\/span><\/p>\n

The horizontal component of speed, vox<\/sub> is always same so vox <\/sub>in part 1 of the projectile motion = vox<\/sub> in part 2 of the projectile motion = 5 m\/s. <\/span><\/p>\n

\"Momentum,Known <\/u><\/span><\/p>\n

vox<\/sub> = 5 m\/s<\/span><\/p>\n

\u03b8 = 60o<\/sup><\/span><\/p>\n

vox <\/sub>and \u03b8 are known so the initial speed (vo<\/sub>) can be calculated.<\/span><\/p>\n

cos \u03b8 = adj \/ hyp<\/span><\/p>\n

cos \u03b8 = vox<\/sub> \/ vo<\/sub><\/span><\/p>\n

vo <\/sub>= vox<\/sub> \/ cos \u03b8 = 5 \/ cos 60o <\/sup>= 5 \/ 0.5 = 10 m\/s<\/span><\/p>\n

The initial speed (vo<\/sub>) is 10 m\/s.<\/span><\/p>\n

The initial speed of the ball <\/i>(v<\/i>o<\/i><\/sub>) <\/i>in projectile motion <\/i>= <\/i>the final speed of the ball <\/i>(v<\/i>t<\/i><\/sub>) <\/i>in the equation of impulse-momentum.<\/i><\/span><\/p>\n

Determine the magnitude of force <\/b><\/u>(F)<\/b><\/u><\/span><\/p>\n

F = 0.2 vt <\/sub> \/ 0.01<\/span><\/p>\n

F = 0.2 (10) \/ 0.01<\/span><\/p>\n

F = 2 \/ 0.01<\/span><\/p>\n

F = 200 Newton <\/span><\/p>\n

    \n
  1. What is the relationship between impulse and change in momentum?<\/strong><\/span>\n