Question

1.) A 200-g ball moving toward the right at a speed of 30 cm/ strikes a 400-g ball moving toward the left at an unknown speed, v. After collision they stick together and remain motionless. Find the speed of the 400-g ball before the collision.

2.) A 5kg mass moving to the right at a speed of 10m/s collides with an unknown mass initially moving to the left at the same speed. After collision they both moved to the right at the speed of 4m/s. Find the unknown mass.

Answer 1

Q1) -15 cm/s and Q2) 2.143 kg

Step-by-step explanation:

Q1) mass of the first ball: M1 = 200 g

initial velocity of the first ball: U1 = 30 cm/s

mass of the second ball: M2 = 400 g

initial velocity of the second ball: U2 = X cm/s

final velocity of the first ball: V1 = 0 cm/s

final velocity of the second ball: V2 = 0 cm/s

By using the law of conservation of momentum that states:

initial momentum = final momentum

(M1 x U1) + (M2 x U2) = (M1 x V1) + (M2 x V2)

however as after collision both the bass stick together their masses combine and the move together in the same direction with the same speed.

therefore;

(M1 x U1) + (M2 x U2) = (M1 + M2)(V)

(200 x 30) + (400 x X) = (400 + 200) (0)

6000 + 400X = 800 x 0

6000 + 400X = 0

X = -6000/400

X = -15 cm/s

( the minus sign indicates the opposite direction i.e. left)

Q2) M1 = 5 kg

M2 = X kg

U1 = 10m/s

U2 = -10 m/s ( as its the opposite direction)

V1 = 4 m/s

V2 = 4m/s

using the law of conservation of momentum:

(M1 x U1) + (M2 x U2) = (M1 x V1) + (M2 x V2)

( 5 x 10) + ( X x (-10)) = (5 x 4) + (X x 4)

50 - 10X = 20 + 4X

50 - 20 = 4X + 10X

30 = 14X

30/14 = X

2.143 kg = X