Question

Let’s begin by calculating the total momentum of two moving objects. Suppose a small compact car with a mass of 1000 kg is traveling north on Morewood Avenue at a speed of 15 m/sm/s. At the intersection of Morewood and Fifth Avenues, it collides with a truck with a mass of 2000 kg that is traveling east on Fifth Avenue at 10 m/sm/s. Treating each vehicle as a particle, find the total momentum (magnitude and direction) just before the collision.

If the car is moving at 18 m/s, how fast must the truck move for the total momentum vector to make a 60 ° angle with the +x axis (and 30° with the +y axis)?

Express the velocity to two significant figures and include the appropriate units.

VB 10 m/s B me = 2000 kg (east) 1 VA = 15 m/s А ma = 1000 kg

Answer 1

Ans :- m. - 1000 kg. = is mis mg = 2000 kg V 10 m/s L momentum of car A before collision is - PA = MAV 15000 kgm/s momentum of truck B before collision is - = more 20000 kgm/s Hence, magnitude of total momentum collision P = J P + Pg² √ (15000)² + (20000) Po before P= 25ooo K8 m/s Direction of total momentum before collision- 15ooo tano = PA = sw PB 2ooo e = tant (3) 36.86 37 with x-axis.

now a velocity 16 m/s. A ſ A since car is moving moving with 18 m/s Let the e speed of truck is u V = voi along a-axis as - - 1000 X 18 = IA 18000 kgm/s cy PB B 2000 V B 0 = 60° with the x-axis. tan 60 = - PA PB 3 18000 2000VA ve 5.196 mis -