QUESTION 3 2.5 points Save Answer Assume momentum is conserved in this case. Car
ID: 3282327 • Letter: Q
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QUESTION 3 2.5 points Save Answer Assume momentum is conserved in this case. Cart A of mass 0.737kg is moving on a leveled track with a speed 0.458m/s. It then collides with cart B of unknown mass. After collision, the two carts are stuck together with a speed 0.246m/s. What is the mass of cart B (in unit of kg)? QUESTION 4 2.5 points Save Answer Assume momentum is conserved in this case. A cart of mass 0.589kg is moving on a leveled track with a speed 0.433m/s. It then collides with another cart of mass 0.534kg, and become stuck with it. What is the speed in (m/s) after collision?Explanation / Answer
Answer 3) Acc. to law of conservation of momentum,
m1u1 + m2u2 =m1v1 + m2v2
Since the initial velocity of second object is zero as it is in rest position, u2=0
Therefore, m1u1 = m1v1 + m2v2
After collision, two bodies stick together and behave as one system, and moves with velocity 'V'
Therfore, m1u1= (m1+m2) V .......................(1)
On putting the values in eq. (1)
m1= 0.737kg, u1=0.458m/s, V=0.246m/s, m2=?
0.737 * 0.485 = (0.737 + m2) 0.246
0.357445 = (0.737 + m2) 0.246
0.357445 / 0.246 = 0.737 + m2
1.4530 = 0.737 + m2
m2 = 1.4530-0.737
therefore, m2 = 0.716 kg
Answer 4) Acc. to law of conservation of momentum,
m1u1 + m2u2 =m1v1 + m2v2
Since the initial velocity of second object is zero as it is in rest position, u2=0
Therefore, m1u1 = m1v1 + m2v2
After collision, two bodies stick together and behave as one system, and moves with velocity 'V'
Therfore, m1u1= (m1+m2) V .......................(1)
On putting the values in eq. (1)
m1= 0.589kg, u1=0.433m/s, V=? , m2=0.534
0.589 * 0.433 = (0.589 + 0.534) V
0.255037 = (1.123) V
0.255037 / 1.123 = V
V= 0.227 m/s
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